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• Set Recurring payment with different initial payment in Paypal

  - by www.sapnaedu.in

  I am developing a payment option in a PHP based application. The user can choose Paypal or Paypal recurring method to make a payment. However, the user would pay $50 for the first time and $40 starting from next month. However, when the user chooses the Paypal recurring option and he pays $50, Paypal automatically chooses $50 from the next month onwards. Is it possible to set the different initial payment and recurring payment ? Here is the part of the code : echo "<input type=\"hidden\" name=\"no_shipping\" value=\"1\"/>\n"; echo "<input type=\"hidden\" name=\"a3\" value=\"".$amt."\"/>\n"; echo "<input type=\"hidden\" name=\"p3\" value=\"1\"/>\n"; echo "<input type=\"hidden\" name=\"t3\" value=\"M\"/>\n"; echo "<input type=\"hidden\" name=\"src\" value=\"1\"/>\n"; echo "<input type=\"hidden\" name=\"sra\" value=\"1\"/>\n"; echo "<input type=\"hidden\" name=\"no_note\" value=\"1\"/>\n"; Thanks Kiran

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• Bus Timetable database design

  - by paddydub

  hi, I'm trying to design a db to store the timetable for 300 different bus routes, Each route has a different number of stops and different times for Monday-Friday, Saturday and Sunday. I've represented the bus departure times for each route as follows, I'm not sure if i should have null values in the table, does this look ok? route,Num,Day, t1, t2, t3, t4 t5 t6 t7 t8 t9 t10 117, 1, Monday, 9:00, 9:30, 10:50, 12:00, 14:00 18:00 19:00 null null null 117, 2, Monday, 9:03, 9:33, 10:53, 12:03, 14:03 18:03 19:03 null null null 117, 3, Monday, 9:06, 9:36, 10:56, 12:06, 14:06 18:06 19:06 null null null 117, 4, Monday, 9:09, 9:39, 10:59, 12:09, 14:09 18:09 19:09 null null null . . . 117, 20, Monday, 9:39, 10.09, 11:39, 12:39, 14:39 18:39 19:39 null null null 119, 1, Monday, 9:00, 9:30, 10:50, 12:00, 14:00 18:00 19:00 20:00 21:00 22:00 119, 2, Monday, 9:03, 9:33, 10:53, 12:03, 14:03 18:03 19:03 20:03 21:03 22:03 119, 3, Monday, 9:06, 9:36, 10:56, 12:06, 14:06 18:06 19:06 20:06 21:06 22:06 119, 4, Monday, 9:09, 9:39, 10:59, 12:09, 14:09 18:09 19:09 20:09 21:09 22:09 . . . 119, 37, Monday, 9:49, 9:59, 11:59, 12:59, 14:59 18:59 19:59 20:59 21:59 22:59 139, 1, Sunday, 9:00, 9:30, 20:00 21:00 22:00 null null null null null 139, 2, Sunday, 9:03, 9:33, 20:03 21:03 22:03 null null null null null 139, 3, Sunday, 9:06, 9:36, 20:06 21:06 22:06 null null null null null 139, 4, Sunday, 9:09, 9:39, 20:09 21:09 22:09 null null null null null . . . 139, 20, Sunday, 9:49, 9:59, 20:59 21:59 22:59 null null null null null

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• How to synchronize placing an object in JNDI across processes?

  - by indra

  I am trying to place an object in JNDI, so that only one of the progam should be able to place it in JNDI. is there any global lock that can be used in J2EE environment. Is RMI can be used for this purpose? please provide any reference links. Thanks in advance. Also, what is NameAlreadyBoundexception? I am trying to use it as a method to synchronize, i.e, only one program places it in JNDI and if other trying to bind should get that exception. But when i am testing the multiple binding I am not getting the Exception.And second binding is done. look up is giving the second object bound. here is my code: private static String JNDI_NAME = "java:comp/env/test/something"; public class TestJNDI { private static String JNDI_NAME = "java:comp/env/test/something"; public static void main(String[] args) throws NamingException { Hashtable env = new Hashtable(); env.put(Context.INITIAL_CONTEXT_FACTORY,"weblogic.jndi.WLInitialContextFactory"); env.put(Context.PROVIDER_URL,"t3://127.0.0.1:7001"); Context ctx = new InitialContext(env); System.out.println("Initial Context created"); String obj1 = "obj1"; String obj2 = "obj2"; try{ ctx.bind(JNDI_NAME, obj1); System.out.println("Bind Sucess"); }catch(NameAlreadyBoundException ne ){ // already bound System.out.println("Name already bound"); } ctx.close(); Context ctx2 = new InitialContext(env); try{ // Second binding to the same name not giving the Exception?? ctx2.bind(JNDI_NAME, obj2); System.out.println("Re Bind Sucess"); }catch(NameAlreadyBoundException ne ){ // already bound System.out.println("Name already bound"); } String lookedUp = (String) ctx2.lookup(JNDI_NAME); System.out.println("LookedUp Object"+lookedUp); ctx2.close(); } }

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• Not getting the right expected output for my Mysql Query?

  - by user1878107

  i've 4 tables as shown below doctors id name ------------ 1 Mathew 2 Praveen 3 Rosie 4 Arjun 5 Denis doctors_appointments id doctors_id patient_name contact date status -------------------------------------------------------------------------------------- 1 5 Nidhin 9876543210 2012-12-10 15:39:41 Registered 2 5 Sunny 9876543210 2012-12-18 15:39:48 Registered 3 5 Mani 9876543210 2012-12-12 15:39:57 Registered 4 2 John 9876543210 2012-12-24 15:40:09 Registered 5 4 Raj 9876543210 2012-12-05 15:41:57 Registered 6 3 Samuel 9876543210 2012-12-14 15:41:33 Registered 7 2 Louis 9876543210 2012-12-24 15:40:23 Registered 8 1 Federick 9876543210 2012-12-28 15:41:05 Registered 9 2 Sam 9876543210 2012-12-12 15:40:38 Registered 10 4 Sita 9876543210 2012-12-12 15:41:00 Registered doctors_dutyplan id doctor_id weeks time no_of_patients ------------------------------------------------------------------ 1 1 3,6,7 9:00am-1:00pm 10 2 2 3,4,5 1:00pm-4:00pm 7 3 3 3,6,7 10:00am-2:00pm 10 4 4 3,4,5,6 8:30am-12:30pm 12 5 5 3,4,5,6,7 9:00am-4:00pm 30 emp_leave id empid leavedate -------------------------------- 1 2 2012-12-05 14:42:36 2 2 2012-12-03 14:42:59 3 3 2012-12-03 14:43:06 4 3 2012-12-06 14:43:14 5 5 2012-12-04 14:43:24 My task is to find all the days in a month in which the doctor is available excluding the leave dates. My query what is wrote is given below: SELECT DATE_ADD( '2012-12-01', INTERVAL ROW DAY ) AS Date, ROW +1 AS DayOfMonth FROM ( SELECT @row := @row +1 AS ROW FROM ( SELECT 0 UNION ALL SELECT 1 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 )t1, ( SELECT 0 UNION ALL SELECT 1 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 )t2, ( SELECT @row := -1 )t3 LIMIT 31 )b WHERE DATE_ADD( '2012-12-01', INTERVAL ROW DAY ) BETWEEN '2012-12-01' AND '2012-12-31' AND DAYOFWEEK( DATE_ADD( '2012-12-01', INTERVAL ROW DAY ) ) =2 AND DATE_ADD( '2012-12-01', INTERVAL ROW DAY ) NOT IN ( SELECT DATE_FORMAT( l.leavedate, '%Y-%m-%d' ) AS date FROM doctors_dutyplan d LEFT JOIN emp_leave AS l ON d.doctor_id = l.empid WHERE doctor_id =2 ) This works fine for all doctors who took any leave in a particular day in a month (here in the example it is Decemeber 2012). and the result is shown below: Date DayOfMonth ----------------------- 2012-12-10 10 2012-12-17 17 2012-12-24 24 2012-12-31 31 But on the other hand for the doctors who did'nt took any leave , for that my query is showing empty table, example for the doctor Mathew whose id is 1, my query returns an empty result can anyone please tell a solution for this problem. Thanks in advance.

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• jQuery DOM manipulation

  - by ufw

  I have different php output in jQuery-based tabs. This output is formed from database and it consists of several <div>'s. Every time I click any tab it sends AJAX request to the php script which performs "SELECT" request and gives result back as response to AJAX request from the tab. $(document).ready(function() { $('ul.tabs li').css('cursor', 'pointer'); $('ul.tabs.tabs1 li').click(function(){ var thisClass = this.className.slice(0,2); $('div.t1').hide(); $('div.t2').hide(); $('div.t3').hide(); $('div.t4').hide(); $('div.' + thisClass).show('fast'); $('ul.tabs.tabs1 li').removeClass('tab-current'); $(this).addClass('tab-current'); var data = thisClass; $.ajax({ type:"GET", url:"handler.php?name="+data, data:data, success:function(html) { $('div.' + thisClass).html(html); } }); return false; }); }); //Server-side PHP script: <?php $name = $_GET[name]; switch ($name) { case "t1": query_and_output_1(); case "t2": query_and_output_2(); // etc... } ?> The problem is that the first tab contains output that must be in second and third ones as well, the second also contains the third tab output. Sorry for such a rubbishy question, I used to work with server side and unfortunately I'm not familiar with DOM and jQuery yet. Thanks.

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• Syntax for documenting JSON structure

  - by Roman A. Taycher

  So I'm trying to document the format of the json returned by an api I am writing against and I'd like to know if there is any popular format for the documentation of json structure. Note I'm not trying to to test or validate anything, I'm just using this for documentation. Also some ways to add comments to non-constants(items always returned w/ the same value) would be nice. This the not totally thought out scheme I'm currently using: Plain names refer to identifiers or types. Some types have type-comment Strings that appear to be constant(always returned for that type of request) strings are "str" Constant Numbers would be just the number Constant null is null Booleans are true/false for constant booleans or Boolean otherwise [a,b,c] are lists with 3 items a,b,c [... ...] is a list of repeating elements of some types/constants/patterns {a:A,b:B,c:c} and {... ...} is the same for a dictionary. example: story := [header,footer] header := {"data":realHeader,"kind":"Listing"} realHeader := {"after": null, "before": null, "children": [{"data": realRealHeader, "kind": "t3"}], "modhash": ""} footer := {"data":AlmostComments,"kind":"Listing"} AlmostComments := {"data": {"after": null, "before": null, "children": comments, "modhash": ""}, "kind": "t1"} comments := [...{"data":comment, "kind":"t1"}...] realRealHeader := {"author": string, "clicked": boolean, "created": int, "created_utc": int, "domain": "code.reddit.com", "downs": int, "hidden": boolean, "id": string-id, "is_self": boolean, "levenshtein": null, "likes": null, "media": null, "media_embed": { }, "name": string-id, "num_comments": int, "over_18": false, "permalink": string-urlLinkToStoryStartingFrom/r, "saved": false, "score": int, "selftext": string, "selftext_html": string-html, "subreddit": string-subredditname, "subreddit_id": string-id, "thumbnail": "", "title": string, "ups": int, "url": "http://code.reddit.com/" } comments := { "author": string, "body": string-body_html-wout-html, "body_html": string-html-formated, "created": int, "created_utc": int, "downs": int, "id": string-id, "levenshtein": null, "likes": null, "link_id": string-id, "name": string-id", "parent_id": string-id, "replies": AlmostComments or null, "subreddit": string-subredditname, "subreddit_id": string-id, "ups": int }

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• jQuery AJAX tabs + PHP

  - by ufw

  Hi, seems like I'm stuck with jQuery tabs. I'm trying to pass selected tab name to some php script but is seems like it doesn't get any data. This is how tabs normally work without any response from server side: http://pastebin.com/KBxj7p5k And this is how I try to pass the the current tab name to the server: $(document).ready(function() { $('ul.tabs li').css('cursor', 'pointer'); $('ul.tabs.tabs1 li').click(function(){ var thisClass = this.className.slice(0,2); $('div.t1').hide(); $('div.t2').hide(); $('div.t3').hide(); $('div.t4').hide(); $('div.' + thisClass).show('fast'); $('ul.tabs.tabs1 li').removeClass('tab-current'); $(this).addClass('tab-current'); var name = thisClass; var data = 'name='+name; $.ajax ({ type:"GET", url:"handler.php", data:data, success:function(html) { thisClass.html(html); } }); }); Thanks.

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• Efficiently select top row for each category in the set

  - by VladV

  I need to select a top row for each category from a known set (somewhat similar to this question). The problem is, how to make this query efficient on the large number of rows. For example, let's create a table that stores temperature recording in several places. CREATE TABLE #t ( placeId int, ts datetime, temp int, PRIMARY KEY (ts, placeId) ) -- insert some sample data SET NOCOUNT ON DECLARE @n int, @ts datetime SELECT @n = 1000, @ts = '2000-01-01' WHILE (@n>0) BEGIN INSERT INTO #t VALUES (@n % 10, @ts, @n % 37) IF (@n % 10 = 0) SET @ts = DATEADD(hour, 1, @ts) SET @n = @n - 1 END Now I need to get the latest recording for each of the places 1, 2, 3. This way is efficient, but doesn't scale well (and looks dirty). SELECT * FROM ( SELECT TOP 1 placeId, temp FROM #t WHERE placeId = 1 ORDER BY ts DESC ) t1 UNION ALL SELECT * FROM ( SELECT TOP 1 placeId, temp FROM #t WHERE placeId = 2 ORDER BY ts DESC ) t2 UNION ALL SELECT * FROM ( SELECT TOP 1 placeId, temp FROM #t WHERE placeId = 3 ORDER BY ts DESC ) t3 The following looks better but works much less efficiently (30% vs 70% according to the optimizer). SELECT placeId, ts, temp FROM ( SELECT placeId, ts, temp, ROW_NUMBER() OVER (PARTITION BY placeId ORDER BY ts DESC) rownum FROM #t WHERE placeId IN (1, 2, 3) ) t WHERE rownum = 1 The problem is, during the latter query execution plan a clustered index scan is performed on #t and 300 rows are retrieved, sorted, numbered, and then filtered, leaving only 3 rows. For the former query three times one row is fetched. Is there a way to perform the query efficiently without lots of unions?

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• Haskell: How to compose `not` with a function of arbitrary arity?

  - by Hynek -Pichi- Vychodil

  When I have some function of type like f :: (Ord a) => a -> a -> Bool f a b = a > b I should like make function which wrap this function with not. e.g. make function like this g :: (Ord a) => a -> a -> Bool g a b = not $ f a b I can make combinator like n f = (\a -> \b -> not $ f a b) But I don't know how. *Main> let n f = (\a -> \b -> not $ f a b) n :: (t -> t1 -> Bool) -> t -> t1 -> Bool Main> :t n f n f :: (Ord t) => t -> t -> Bool *Main> let g = n f g :: () -> () -> Bool What am I doing wrong? And bonus question how I can do this for function with more and lest parameters e.g. t -> Bool t -> t1 -> Bool t -> t1 -> t2 -> Bool t -> t1 -> t2 -> t3 -> Bool

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• restrict duplicate rows in specific columns in mysql

  - by JPro

  I have a query like this : select testset, count(distinct results.TestCase) as runs, Sum(case when Verdict = "PASS" then 1 else 0 end) as pass, Sum(case when Verdict <> "PASS" then 1 else 0 end) as fails, Sum(case when latest_issue <> "NULL" then 1 else 0 end) as issues, Sum(case when latest_issue <> "NULL" and issue_type = "TC" then 1 else 0 end) as TC_issues from results join testcases on results.TestCase = testcases.TestCase where platform = "T1_PLATFORM" AND testcases.CaseType = "M2" and testcases.dummy <> "flag_1" group by testset order by results.TestCase The result set I get is : testset runs pass fails issues TC_issues T1 66 125 73 38 33 T2 18 19 16 16 15 T3 57 58 55 55 29 T4 52 43 12 0 0 T5 193 223 265 130 22 T6 23 12 11 0 0 My problem is, this is a result table which has testcases running multiple times. So, I am able to restrict the runs using the distinct TestCases but when I want the pass and fails, since I am using case I am unable to eliminate the duplicates. Is there any way to achieve what I want? any help please? thanks.

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• When to use CTEs to encapsulate sub-results, and when to let the RDBMS worry about massive joins.

  - by IanC

  This is a SQL theory question. I can provide an example, but I don't think it's needed to make my point. Anyone experienced with SQL will immediately know what I'm talking about. Usually we use joins to minimize the number of records due to matching the left and right rows. However, under certain conditions, joining tables cause a multiplication of results where the result is all permutations of the left and right records. I have a database which has 3 or 4 such joins. This turns what would be a few records into a multitude. My concern is that the tables will be large in production, so the number of these joined rows will be immense. Further, heavy math is performed on each row, and the idea of performing math on duplicate rows is enough to make anyone shudder. I have two questions. The first is, is this something I should care about, or will SQL Server intelligently realize these rows are all duplicates and optimize all processing accordingly? The second is, is there any advantage to grouping each part of the query so as to get only the distinct values going into the next part of the query, using something like: WITH t1 AS ( SELECT DISTINCT... [or GROUP BY] ), t2 AS ( SELECT DISTINCT... ), t3 AS ( SELECT DISTINCT... ) SELECT... I have often seen the use of DISTINCT applied to subqueries. There is obviously a reason for doing this. However, I'm talking about something a little different and perhaps more subtle and tricky.

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• Certain SFTP user cannot connect

  - by trobrock

  I have my Ubuntu Server set up so users with the group of sftponly can connect with sftp, but have a shell of /bin/false, and they connect to their home directories. This is working fine with three of the user accounts I have. But I added a new user account today the same way that I added the others and it will not successfully connect. sftp -vvv user@hostname debug1: Next authentication method: password user@hostname's password: debug3: packet_send2: adding 48 (len 73 padlen 7 extra_pad 64) debug2: we sent a password packet, wait for reply debug1: Authentication succeeded (password). debug2: fd 5 setting O_NONBLOCK debug3: fd 6 is O_NONBLOCK debug1: channel 0: new [client-session] debug3: ssh_session2_open: channel_new: 0 debug2: channel 0: send open debug1: Requesting [email protected] debug1: Entering interactive session. debug1: channel 0: free: client-session, nchannels 1 debug3: channel 0: status: The following connections are open: #0 client-session (t3 r-1 i0/0 o0/0 fd 5/6 cfd -1) debug3: channel 0: close_fds r 5 w 6 e 7 c -1 debug1: fd 0 clearing O_NONBLOCK debug3: fd 1 is not O_NONBLOCK Connection to hostname closed by remote host. Transferred: sent 2176, received 1848 bytes, in 0.0 seconds Bytes per second: sent 127453.3, received 108241.6 debug1: Exit status -1 Connection closed For a successful user: sftp -vvv good_user@hostname debug1: Next authentication method: password good_user@hostname's password: debug3: packet_send2: adding 48 (len 63 padlen 17 extra_pad 64) debug2: we sent a password packet, wait for reply debug1: Authentication succeeded (password). debug2: fd 5 setting O_NONBLOCK debug3: fd 6 is O_NONBLOCK debug1: channel 0: new [client-session] debug3: ssh_session2_open: channel_new: 0 debug2: channel 0: send open debug1: Requesting [email protected] debug1: Entering interactive session. debug2: callback start debug2: client_session2_setup: id 0 debug1: Sending subsystem: sftp debug2: channel 0: request subsystem confirm 1 debug2: fd 3 setting TCP_NODELAY debug2: callback done debug2: channel 0: open confirm rwindow 0 rmax 32768 debug2: channel 0: rcvd adjust 2097152 debug2: channel_input_status_confirm: type 99 id 0 debug2: subsystem request accepted on channel 0 debug2: Remote version: 3 debug2: Server supports extension "[email protected]" revision 1 debug2: Server supports extension "[email protected]" revision 2 debug2: Server supports extension "[email protected]" revision 2 debug3: Sent message fd 3 T:16 I:1 debug3: SSH_FXP_REALPATH . -> / sftp> I cannot figure out why one user will work and the other wont, I have restart the ssh service after adding the user. I have even removed the user and added them again to be sure I am adding it correctly.

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• printing reverse in singly link list using pointers

  - by theoneabhinav

  i have been trying this code. i think the logic is ok but the program terminates abruptly when the display_rev function is called here is code of display_rev void display_rev(emp_node *head) { emp_node *p=head, *q; while(p->next != NULL) p=p->next; while(p!=head || p==head){ q=head; display_rec(p); while(q->next != p) q=q->next; p=q; } } here is my whole code #include<stdio.h> #include<stdlib.h> #include<ctype.h> #include<string.h> //Declarations=============================================================== typedef struct //employee record { int emp_id; char name[150]; char mob_no[11]; float salary; int proj[5]; struct emp_node *next; } emp_node; emp_node* add_rec(emp_node*); emp_node* create_db(emp_node*); emp_node* search_db(emp_node*, int); emp_node* delete_rec(emp_node*, int); void read_name(emp_node*); void read_mob(emp_node*); void display_db(emp_node*); void display_rec(emp_node*); void display_rev(emp_node*); void modify_rec(emp_node*); void swtch(emp_node*); //=========================================================================== int main() { char ans; emp_node *head = NULL; head = create_db(head); display_db(head); do { swtch(head); printf("\n\n\tDo you want to continue (y/n) : "); getchar(); scanf("%c", &ans); } while (ans == 'y' || ans == 'Y'); return 0; } //Definitions================================================================ emp_node* create_db(emp_node *head) //database creation { int i = 1, no; emp_node *p; printf("Enter number of employees:"); scanf("%d", &no); printf("\n\n"); head = (emp_node *) malloc(sizeof(emp_node)); head = add_rec(head); head->next = NULL; p = head; while (i < no) { p->next = (emp_node *) malloc(sizeof(emp_node)); p = p->next; p = add_rec(p); p->next = NULL; i++; } return head; } emp_node* add_rec(emp_node *p) //new record { int j; printf("\n\tEmployee ID : "); scanf("%d", &(p->emp_id)); printf("\n\tFirst Name:"); read_name(p); printf("\n\tMobile No.:"); read_mob(p); printf("\n\tSalary :"); scanf("%f", &(p->salary)); printf( "\n\tEnter \"1\" for the projects employee is working on, otherwise enter \"0\": \n"); for (j = 0; j < 5; j++) { printf("\n\t\tProject No. %d : ", j + 1); scanf("%d", &(p->proj[j])); while (p->proj[j] != 1 && p->proj[j] != 0) { printf("\n\nInvalid entry!! Please re-enter."); printf("\n\t\tProject No. %d : ", j + 1); scanf("%d", &(p->proj[j])); } } printf("\n\n\n"); return p; } void read_name(emp_node *p) //validation for name { int j, len; scanf("%s", p->name); len = strlen(p->name); for (j = 0; j < len; j++) { if (!isalpha(p->name[j])) { printf( "\n\n\tInvalid name!!Can contain only characters. Please Re-enter.\n"); printf("\n\tName : "); read_name(p); } } } void read_mob(emp_node *p) //validation for mobile no. { int j; scanf("%s", p->mob_no); while (strlen(p->mob_no) != 10) { printf("\n\nInvalid Mobile No!!Please Re-enter"); printf("\n\n\tMobile No.:"); read_mob(p); } for (j = 0; j < 10; j++) { if (!(48 <= p->mob_no[j] && p->mob_no[j] <= 57)) { printf( "\n\nInvalid Mobile No!!Can contain only digits. Please Re-enter."); printf("\n\n\tMobile No.:"); read_mob(p); } } } void display_db(emp_node *head) //displaying whole database { emp_node *p; p = head; printf("\n\n\t\t****** EMPLOYEE DATABASE ******\n"); printf( "\n=============================================================================="); printf("\n Id.\t Name\t\t Mobile No\t Salary\t Projects\n"); while (p != NULL) { display_rec(p); p = p->next; printf("\n\n\n"); } printf( "\n=============================================================================="); } void swtch(emp_node *head) //function for menu and switch case { int cho, x; emp_node *p; printf("\n\n\t\t****** MENU ******"); printf( "\n\n\t1. insert Record\n\t2. Search Record\n\t3. Modify Record\n\t4. Delete Record\n\t5. Display Reverse\n\t6. Exit"); printf("\n\tWhich operation do you want to perform? "); scanf("%d", &cho); switch (cho) { case 1: p=head; while(p->next != NULL) p=p->next; p->next = (emp_node *) malloc(sizeof(emp_node)); p=p->next; p = add_rec(p); p->next = NULL; display_db(head); break; case 2: printf("\n\n\tEnter employee ID whose record is to be Searched :"); scanf("%d", &x); p = search_db(head, x); if (p == NULL) printf("\n\nRecord not found."); else display_rec(p); break; case 3: printf("\n\n\tEnter employee ID whose record is to be modified :"); scanf("%d", &x); p = search_db(head, x); if (p == NULL) printf("\n\nRecord not found."); else modify_rec(p); display_db(head); break; case 4: printf("\n\n\tEnter employee ID whose record is to be deleted :"); scanf("%d", &x); head = delete_rec(head, x); display_db(head); break; case 5: display_rev(head); case 6: exit(0); default: printf("Invalid Choice!! Please try again."); } } emp_node* search_db(emp_node *head, int id) //search database { emp_node *p = head; while (p != NULL) { if (p->emp_id == id) return p; p = p->next; } return NULL; } void display_rec(emp_node *p) //display a single record { int j; printf("\n %d", p->emp_id); printf("\t %10s", p->name); printf("\t %10s", p->mob_no); printf("\t %05.2f", p->salary); printf("\t "); for (j = 0; j < 5; j++) { if (p->proj[j] == 1) printf(" %d,", j + 1); } } void modify_rec(emp_node *p) //modifying a record { int j, cho; char ch1, edt; do { printf( "\n\t1. Name\n\t2. Email Address\n\t3. Mobile No.\n\t4. Salary\n\t5. Date of birth\n\t6. Projects\n"); printf("Enter your choice : "); scanf("%d", &cho); switch (cho) { case 1: printf("\n\tPrevious name:%s", p->name); printf("\n\tDo you want to edit ? (y/n)"); getchar(); scanf("%c", &ch1); if (ch1 == 'y' || ch1 == 'Y') { printf("\n\tEnter New Name:"); read_name(p); } break; case 2: printf("\n\tPrevious Mobile No. : %s", p->mob_no); printf("\n\tDo you want to edit ? (y/n)"); getchar(); scanf("%c", &ch1); if (ch1 == 'y' || ch1 == 'Y') { printf("\n\tEnter New Mobile No. :"); read_mob(p); } break; case 3: printf("\n\tPrevious salary is : %f", p->salary); printf("\n\tDo you want to edit ? (y/n)"); getchar(); scanf("%c", &ch1); if (ch1 == 'y' || ch1 == 'Y') { printf("\n\tEnter New salary:"); scanf("%f", &(p->salary)); } break; case 4: printf("the employee is currently working on project no. "); for (j = 0; j < 5; j++) { if (p->proj[j] == 1) printf(" %d,", j + 1); } printf("\n\tDo you want to edit ? (y/n)"); getchar(); scanf("%c", &ch1); if (ch1 == 'y' || ch1 == 'Y') { printf( "\n\tEnter \"1\" for the projects employee is working on : \n"); for (j = 0; j < 5; j++) { printf("\n\t\tProject No. %d : ", j + 1); scanf("%d", &(p->proj[j])); while (p->proj[j] != 1) { printf("\n\nInvalid entry!! Please re-enter."); printf("\n\t\tProject No. %d : ", j + 1); scanf("%d", &(p->proj[j])); } } } break; default: printf("\n\nInvalid Choice!! Please Try again."); } printf("\n\nDo you want to edit any other fields ?(y/n)"); getchar(); scanf("%c", &edt); } while (edt == 'y' || edt == 'Y'); } emp_node* delete_rec(emp_node *head, int id) //physical deletion of record { emp_node *p = head, *q; if (head->emp_id == id) { head = head->next; free(p); return head; } else { q = p->next; while (q->emp_id != id) { p = p->next; q = q->next; } if (q->next == NULL) p->next = NULL; else p->next = q->next; free(q); return head; } } void display_rev(emp_node *head) { emp_node *p=head, *q; while(p->next != NULL) p=p->next; while(p!=head || p==head){ q=head; display_rec(p); while(q->next != p) q=q->next; p=q; } }

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• Hello With Oracle Identity Manager Architecture

  - by mustafakaya

  Hi, my name is Mustafa! I'm a Senior Consultant in Fusion Middleware Team and living in Istanbul,Turkey. I worked many various Java based software development projects such as end-to-end web applications, CRM , Telco VAS and integration projects.I want to share my experiences and research about Fusion Middleware Products in this column. Customer always wants best solution from software consultants or developers. Solution will be a code snippet or change complete architecture. We faced different requests according to the case of customer. In my posts i want to discuss Fusion Middleware Products Architecture or how can extend usability with apis or UI customization and more and I look forward to engaging with you on your experiences and thoughts on this.  In my first post, i will be discussing Oracle Identity Manager architecture  and i plan to discuss Oracle Identity Manager 11g features in next posts. Oracle Identity Manager System Architecture Oracle Identity Governance includes Oracle Identity Manager,Oracle Identity Analytics and Oracle Privileged Account Manager. I will discuss Oracle Identity Manager architecture in this post.  In basically, Oracle Identity Manager is a n-tier standard  Java EE application that is deployed on Oracle WebLogic Server and uses  a database .  Oracle Identity Manager presentation tier has three different screen and two different client. Identity Self Service and Identity System Administration are web-based thin client. Design Console is a Java Swing Client that communicates directly with the Business Service Tier.  Identity Self Service provides end-user operations and delegated administration features. System Administration provides system administration functions. And Design Console mostly use for development management operations such as  create and manage adapter and process form,notification , workflow desing, reconciliation rules etc. Business service tier is implemented as an Enterprise JavaBeans(EJB) application. So you can extense Oracle Identity Manager capabilities.  -The SMPL and EJB APIs allow develop custom plug-ins such as management roles or identities.  -Identity Services allow use core business capabilites of Oracle Identity Manager such as The User provisioning or reconciliation service. -Integration Services allow develop custom connectors or adapters for various deployment needs. -Platform Services allow use Entitlement Servers, Scheduler or SOA composites. The Middleware tier allows you using capabilites ADF Faces,SOA Suites, Scheduler, Entitlement Server and BI Publisher Reports. So OIM allows you to configure workflows uses Oracle SOA Suite or define authorization policies use with Oracle Entitlement Server. Also you can customization of OIM UI without need to write code and using ADF Business Editor  you can extend custom attributes to user,role,catalog and other objects. Data tiers; Oracle Identity Manager is driven by data and metadata which provides flexibility and adaptability to Oracle Identity Manager functionlities.  -Database has five schemas these are OIM,SOA,MDS,OPSS and OES. Oracle Identity Manager uses database to store runtime and configuration data. And all of entity, transactional and audit datas are stored in database. -Metadata Store; customizations and personalizations are stored in file-based repository or database-based repository.And Oracle Identity Manager architecture,the metadata is in Oracle Identity Manager database to take advantage of some of the advanced performance and availability features that this mode provides. -Identity Store; Oracle Identity Manager provides the ability to integrate an LDAP-based identity store into Oracle Identity Manager architecture.  Oracle Identity Manager uses the human workflow module of Oracle Service Oriented Architecture Suite. OIM connects to SOA using the T3 URL which is front-end URL for the SOA server.Oracle Identity Manager uses embedded Oracle Entitlement Server for authorization checks in OIM engine.  Several Oracle Identity Manager modules use JMS queues. Each queue is processed by a separate Message Driven Bean (MDB), which is also part of the Oracle Identity Manager application. Message producers are also part of the Oracle Identity Manager application. Oracle Identity Manager uses a scheduled jobs for some activities in the background.Some of scheduled jobs come with Out-Of-Box such as the disable users after the end date of the users or you can define your custom schedule jobs with Oracle Identity Manager APIs. You can use Oracle BI Publisher for reporting Oracle Identity Manager transactions or audit data which are in database. About me: Mustafa Kaya is a Senior Consultant in Oracle Fusion Middleware Team, living in Istanbul. Before coming to Oracle, he worked in teams developing web applications and backend services at a telco company. He is a Java technology enthusiast, software engineer and addicted to learn new technologies,develop new ideas. Follow Mustafa on Twitter,Connect on LinkedIn, and visit his site for Oracle Fusion Middleware related tips.

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• Having a hard time having consecutive animations for an attack

  - by Kelby Styler

  So I've been trying to figure this out for about 8 hours now...It's driving me nuts because I am pretty sure that it is something dead simple that I am just not understanding. I had everything working fine when I was just cycling through the animation: Idle - Attack - Attack 1 - Attack 2. Just in an infinite loop. The problem now is that I want it to go Attack - check if x time passes if ctrl pressed before x passes move to Attack 1, if not move back to Idle - Then either Attack 1 or Idle depending on how long has passed. I've almost gotten it a few time, but something always happens where it falls apart if I press ctrl too fast or after multiple cycles of the animation. Any help would be appreciated, I'm just at my wits end on this one. I've been looking at this so long that I just don't know where to go anymore. Code is below, here is the controller using UnityEngine; using System.Collections; public class MeleeAttack : MonoBehaviour { public int damage; public bool Attack; public bool Attack1; public bool Attack2; public bool Idle; private Animator animator; private int attnum = 0; private float count = 2f; private float timeLeft; //Gives value to damage output void MAttackDmg () { if (Input.GetKeyDown (KeyCode.RightControl) || Input.GetKeyDown (KeyCode.LeftControl)) { switch (attnum) { case (0): Attack = true; damage = 2; animator.SetBool ("Attack", Attack); attnum++; Idle = false; animator.SetBool ("Idle", Idle); timeLeft = count; break; case (1): Attack1 = true; damage = 2; animator.SetBool ("Attack1", Attack1); attnum++; Idle = false; animator.SetBool ("Idle", Idle); timeLeft = count; break; case (2): Attack2 = true; damage = 2; animator.SetBool ("Attack2", Attack2); attnum = 0; Idle = false; animator.SetBool ("Idle", Idle); timeLeft = count; break; } } if (Input.GetKeyUp (KeyCode.RightControl) || Input.GetKeyUp (KeyCode.LeftControl)) { switch (attnum) { case (0): Debug.Log ("false"); damage = 0; if (timeLeft <= 0f) { Attack2 = false; animator.SetBool ("Attack2", Attack2); Debug.Log ("t1"); Idle = true; animator.SetBool ("Idle", Idle); attnum = 0; timeLeft = count; } break; case (1): Debug.Log ("false1"); damage = 0; if (timeLeft <= 0f) { Debug.Log ("t2"); Attack = false; animator.SetBool ("Attack", Attack); Idle = true; animator.SetBool ("Idle", Idle); attnum = 0; timeLeft = count; } break; case (2): Debug.Log ("false2"); damage = 0; if (timeLeft <= 0f) { Attack1 = false; animator.SetBool ("Attack1", Attack1); Debug.Log ("t3"); Idle = true; animator.SetBool ("Idle", Idle); attnum = 0; timeLeft = count; } break; } } } // Use this for initialization void Awake () { animator = GetComponent<Animator> (); } // Update is called once per frame void Update () { timeLeft -= Time.deltaTime;; MAttackDmg (); } void Start (){ timeLeft = count; } }

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• OpenSSL Versions in Solaris

  - by darrenm

  Those of you have have installed Solaris 11 or have read some of the blogs by my colleagues will have noticed Solaris 11 includes OpenSSL 1.0.0, this is a different version to what we have in Solaris 10.  I hope the following explains why that is and how it fits with the expectations on binary compatibility between Solaris releases. Solaris 10 was the first release where we included OpenSSL libraries and headers (part of it was actually statically linked into the SSH client/server in Solaris 9).  At time we were building and releasing Solaris 10 the current train of OpenSSL was 0.9.7.  The OpenSSL libraries at that time were known to not always be completely API and ABI (binary) compatible between releases (some times even in the lettered patch releases) though mostly if you stuck with the documented high level APIs you would be fine.   For this reason OpenSSL was classified as a 'Volatile' interface and in Solaris 10 Volatile interfaces were not part of the default library search path which is why the OpenSSL libraries live in /usr/sfw/lib on Solaris 10.  Okay, but what does Volatile mean ? Quoting from the attributes(5) man page description of Volatile (which was called External in older taxonomy): Volatile interfaces can change at any time and for any reason. The Volatile interface stability level allows Sun pro- ducts to quickly track a fluid, rapidly evolving specif- ication. In many cases, this is preferred to providing additional stability to the interface, as it may better meet the expectations of the consumer. The most common application of this taxonomy level is to interfaces that are controlled by a body other than Sun, but unlike specifications controlled by standards bodies or Free or Open Source Software (FOSS) communities which value interface compatibility, it can not be asserted that an incompatible change to the interface specifica- tion would be exceedingly rare. It may also be applied to FOSS controlled software where it is deemed more important to track the community with minimal latency than to provide stability to our customers. It also common to apply the Volatile classification level to interfaces in the process of being defined by trusted or widely accepted organization. These are generically referred to as draft standards. An "IETF Internet draft" is a well understood example of a specification under development. Volatile can also be applied to experimental interfaces. No assertion is made regarding either source or binary compatibility of Volatile interfaces between any two releases, including patches. Applications containing these interfaces might fail to function properly in any future release. Note that last paragraph!  OpenSSL is only one example of the many interfaces in Solaris that are classified as Volatile.  At the other end of the scale we have Committed (Stable in Solaris 10 terminology) interfaces, these include things like the POSIX APIs or Solaris specific APIs that we have no intention of changing in an incompatible way.  There are also Private interfaces and things we declare as Not-an-Interface (eg command output not intended for scripting against only to be read by humans). Even if we had declared OpenSSL as a Committed/Stable interface in Solaris 10 there are allowed exceptions, again quoting from attributes(5): 4. An interface specification which isn't controlled by Sun has been changed incompatibly and the vast majority of interface consumers expect the newer interface. 5. Not making the incompatible change would be incomprehensible to our customers. In our opinion and that of our large and small customers keeping up with the OpenSSL community is important, and certainly both of the above cases apply. Our policy for dealing with OpenSSL on Solaris 10 was to stay at 0.9.7 and add fixes for security vulnerabilities (the version string includes the CVE numbers of fixed vulnerabilities relevant to that release train).  The last release of OpenSSL 0.9.7 delivered by the upstream community was more than 4 years ago in Feb 2007. Now lets roll forward to just before the release of Solaris 11 Express in 2010. By that point in time the current OpenSSL release was 0.9.8 with the 1.0.0 release known to be coming soon.  Two significant changes to OpenSSL were made between Solaris 10 and Solaris 11 Express.  First in Solaris 11 Express (and Solaris 11) we removed the requirement that Volatile libraries be placed in /usr/sfw/lib, that means OpenSSL is now in /usr/lib, secondly we upgraded it to the then current version stream of OpenSSL (0.9.8) as was expected by our customers. In between Solaris 11 Express in 2010 and the release of Solaris 11 in 2011 the OpenSSL community released version 1.0.0.  This was a huge milestone for a long standing and highly respected open source project.  It would have been highly negligent of Solaris not to include OpenSSL 1.0.0e in the Solaris 11 release. It is the latest best supported and best performing version.     In fact Solaris 11 isn't 'just' OpenSSL 1.0.0 but we have added our SPARC T4 engine and the AES-NI engine to support the on chip crypto acceleration. This gives us 4.3x better AES performance than OpenSSL 0.9.8 running on AIX on an IBM POWER7. We are now working with the OpenSSL community to determine how best to integrate the SPARC T4 changes into the mainline OpenSSL.  The OpenSSL 'pkcs11' engine we delivered in Solaris 10 to support the CA-6000 card and the SPARC T1/T2/T3 hardware is still included in Solaris 11. When OpenSSL 1.0.1 and 1.1.0 come out we will asses what is best for Solaris customers. It might be upgrade or it might be parallel delivery of more than one version stream.  At this time Solaris 11 still classifies OpenSSL as a Volatile interface, it is our hope that we will be able at some point in a future release to give it a higher interface stability level. Happy crypting! and thank-you OpenSSL community for all the work you have done that helps Solaris.

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• Updating a database connection password using a script

  - by Tim Dexter

  An interesting customer requirement that I thought was worthy of sharing today. Thanks to James for the requirement and Bryan for the proposed solution and me for testing the solution and proving it works :0) A customers implementation of Sarbanes Oxley requires them to change all database account passwords every 90 days. This is scripted leveraging shell scripts today for most of their environments. But how can they manage the BI Publisher connections? Now, the customer is running 11g and therefore using weblogic on the middle tier, which is the first clue to Bryans proposed solution. To paraphrase and embellish Bryan's solution a little; why not use a JNDI connection from BIP to the database. Then employ the web logic scripting engine to make updates to the JNDI as needed? BIP is completely uninvolved and with a little 'timing' users will be completely unaware of the password updates i.e. change the password when reports are not being executed. Perfect! James immediately tracked down the WLST script that could be used here, http://middlewaremagic.com/weblogic/?p=4261 (thanks Ravish) Now it was just a case of testing the theory. Some steps: Create the JNDI connection in WLS Create the JNDI connection in BI Publisher pointing to the WLS connection Build new data models using or re-point data sources to use the JNDI connection. Create the WLST script to update the WLS JNDI password as needed. Test! Some details. Creating the JNDI connection in web logic is pretty straightforward. Log into hte console and look for Data Sources under the Services section of the home page and click it Click New >> Generic Datasource Give the connection a name. For the JNDI name, prefix it with 'jdbc/' so I have 'jdbc/localdb' - this name is important you'll need it on the BIP side. Select your db type - this will influence the drivers and information needed on the next page. Being a company man, Im using an Oracle db. Click Next Select the driver of choice, theres lots I know, you can read about them I just chose 'Oracle's Driver (Thin) for Instance connections; Versions 9.0.1 and later' Click Next >> Next Fill out the db name (SID), server, port, username to connect and password >> Next Test the config to ensure you can connect. >> Next Now you need to deploy the connection to your BI server, select it and click Next. You're done with the JNDI config. Creating the JNDI connection on the Publisher side is covered here. Just remember to the connection name you created in WLS e.g. 'jdbc/localdb' Not gonna tell you how to do this, go read the user guide :0) Suffice to say, it works. This requires a little reading around the subject to understand the scripting engine and how to execute scripts. Nicely covered here. However a bit of googlin' and I found an even easier way of running the script. ${ServerHome}/common/bin/wlst.sh updatepwd.py Where updatepwd.py is my script file, it can be in another directory. As part of the wlst.sh script your environment is set up for you so its very simple to execute. The nitty gritty: Need to take Ravish's script above and create a file with a .py extension. Its going to need some modification, as he explains on the web page, to make it work in your environment. I played around with it for a while but kept running into errors. The script as is, tries to loop through all of your connections and modify the user and passwords for each. Not quite what we are looking for. Remember our requirement is to just update the password for a given connection. I also found another issue with the script. WLS 10.x does not allow updates to passwords using clear type ie un-encrypted text while the server is in production mode. Its a bit much to set it back to developer mode bounce it, change the passwords and then bounce and then change back to production and bounce again. After lots of messing about I finally came up with the following: ############################################################################# # # Update password for JNDI connections # ############################################################################# print("*** Trying to Connect.... *****") connect('weblogic','welcome1','t3://localhost:7001') print("*** Connected *****") edit() startEdit() print ("*** Encrypt the password ***") en = encrypt('hr') print "Encrypted pwd: ", en print ("*** Changing pwd for LocalDB ***") dsName = 'LocalDB' print 'Changing Password for DataSource ', dsName cd('/JDBCSystemResources/'+dsName+'/JDBCResource/'+dsName+'/JDBCDriverParams/'+dsName) set('PasswordEncrypted',en) save() activate() Its pretty simple and you can expand on it to loop through the data sources and change each as needed. I have hardcoded the password into the file but you can pass it as a parameter as needed using the properties file method. Im not going to get into the detail of that here but its covered with an example here. Couple of points to note: 1. The change to the password requires a server bounce to get the changes picked up. You can add that to the shell script you will use to call the script above. 2. The script above needs to be run from the MW_HOME\user_projects\domains\bifoundation_domain directory to get the encryption libraries set correctly. My command to run the whole script was: d:\oracle\bi_mw\wlserver_10.3\common\bin\wlst.cmd updatepwd.py - where wlst.cmd is the scripting command line and updatepwd.py was my update password script above. I have not quite spoon fed everything you need to make it a robust script but at least you know you can do it and you can work out the rest I think :0)

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• About Solaris 11 and UltraSPARC II/III/IV/IV+

  - by nospam(at)example.com (Joerg Moellenkamp)

  I know that I will get the usual amount of comments like "Oh, Jörg ? you can't be negative about Oracle" for this article. However as usual I want to explain the logic behind my reasoning. Yes ? I know that there is a lot of UltraSPARC III, IV and IV+ gear out there. But there are some very basic questions: Does your application you are currently running on this gear stops running just because you can't run Solaris 11 on it? What is the need to upgrade a system already in production to Solaris 11? I have the impression, that some people think that the systems get useless in the moment Oracle releases Solaris 11. I know that Sun sold UltraSPARC IV+ systems until 2009. The Sun SF490 introduced 2004 for example, that was a Sun SF480 with UltraSPARC IV and later with UltraSPARC IV+. And yes, Sun made some speedbumps. At that time the systems of the UltraSPARC III to IV+ generations were supported on Solaris 8, on Solaris 9 and on Solaris 10. However from my perspective we sold them to customers, which weren't able to migrate to Solaris 10 because they used applications not supported on Solaris 9 or who just didn't wanted to migrate to Solaris 10. Believe it or not ? I personally know two customers that migrated core systems to Solaris 10 in ? well 2008/9. This was especially true when the M3000 was announced in 2008 when it closed the darned single socket gap. It may be different at you site, however that's what I remember about that time when talking with customers. At first: Just because there is no Solaris 11 for UltraSPARC III, IV and IV+, it doesn't mean that Solaris 10 will go away anytime soon. I just want to point you to "Expect Lifetime Support - Hardware and Operating Systems". It states about Premier Support:Maintenance and software upgrades are included for Oracle operating systems and Oracle VM for a minimum of eight years from the general availability date.GA for Solaris 10 was in 2005. Plus 8 years ? 2013 ? at minimum. Then you can still opt for 3 years of "Extended Support" ? 2016 ? at minimum. 2016 your systems purchased in 2009 are 7 years old. Even on systems purchased at the very end of the lifetime of that system generation. That are the rules as written in the linked document. I said minimum The actual dates are even further in the future: Premier Support for Solaris 10 ends in 2015, Extended support ends 2018. Sustaining support ? indefinite. You will find this in the document "Oracle Lifetime Support Policy: Oracle Hardware and Operating Systems".So I don't understand when some people write, that Oracle is less protective about hardware investments than Sun. And for hardware it's the same as with Sun: Service 5 years after EOL as part of Premier Support. I would like to write about a different perspective as well: I have to be a little cautious here, because this is going in the roadmap area, so I will mention the public sources here: John Fowler told last year that we have to expect at at least 3x the single thread performance of T3 for T4. We have 8 cores in T4, as stated by Rick Hetherington. Let's assume for a moment that a T4 core will have the performance of a UltraSPARC core (just to simplify math and not to disclosing anything about the performance, all existing SPARC cores are considered equal). So given this pieces of information, you could consolidate 8 V215, 4 or 8 V245, 2 full blown V445,2 full blown 490, 2 full blown M3000 on a single T4 SPARC processor. The Fowler roadmap prezo talked about 4-socket systems with T4. So 32 V215, 16 to 8 V245, 8 fullblown V445, 8 full blown V490, 8 full blown M3000 in a system image. I think you get the idea. That said, most of the systems we are talking about have already amortized and perhaps it's just time to invest in new systems to yield other advantages like reduced space consumptions, like reduced power consumption, like some of the neat features sun4v gives you, and yes ? reduced number of processor licenses for Oracle and less money for Oracle HW/SW support. As much as I dislike it myself that my own UltraSPARC III and UltraSPARC II based systems won't run on Solaris 11 (and I have quite a few of them in my personal lab), I really think that the impact on production environments will be much less than most people think now. By the way: The reason for this move is a quite significant new feature. I will tell you that it was this feature, when it's out. I assume, telling just a word more could lead to much more time to blog.

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• implementation of text editor in shell programming

  - by Arka Ghosh

  i have made a text editor in C. when i am changing the extension of that file from .c to .sh and compiling the file in the terminal,some error is shown,like for the global variables an external error is shown,and for the functions i have declared errors are shown there also.please help me to solve this. I am sending my code.. include include include int i,j,ec,fg,ec2; char fn[20],e,c,d; FILE *fp1,*fp2,fp; void Create(); void Append(); void Delete(); void Display(); int main() { do { printf("\n\t\t** TEXT EDITOR *"); printf("\n\n\tMENU:\n\t..\n "); printf("\n\t1.CREATE\n\t2.DISPLAY\n\t3.APPEND\n\t4.DELETE\n\t5.EXIT\n"); printf("\n\tEnter your choice: "); scanf("%d",&ec); switch(ec) { case 1: Create(); break; case 2: Display(); break; case 3: Append(); break; case 4: Delete(); break; case 5: exit(1); } }while(1); } void Create() { fp1=fopen("temp.txt","w"); printf("\n\tEnter the text and press '.' to save\n\n\t"); while(1) { c=getchar(); fputc(c,fp1); if(c == '.') { fclose(fp1); printf("\n\tEnter then new filename: "); scanf("%s",fn); fp1=fopen("temp.txt","r"); fp2=fopen(fn,"w"); while(!feof(fp1)) { c=getc(fp1); putc(c,fp2); } fclose(fp2); break; }} } void Display() { printf("\n\tEnter the file name: "); scanf("%s",fn); fp1=fopen(fn,"r"); if(fp1==NULL) { printf("\n\tFile not found!"); goto end1; } while(!feof(fp1)) { c=getc(fp1); printf("%c",c); } end1: fclose(fp1); printf("\n\n\tPress any key to continue.."); } void Delete() { printf("\n\tEnter the file name: "); scanf("%s",fn); fp1=fopen(fn,"r"); if(fp1==NULL) { printf("\n\tFile not found!"); goto end2; } fclose(fp1); if(remove(fn)==0) { printf("\n\n\tFile has been deleted successfully!"); goto end2; } else printf("\n\tError!\n"); end2: printf("\n\n\tPress any key to continue.."); getchar(); } void Append() { printf("\n\tEnter the file name: "); scanf("%s",fn); fp1=fopen(fn,"r"); if(fp1==NULL) { printf("\n\tFile not found!"); goto end3; } while(!feof(fp1)) { c=getc(fp1); printf("%c",c); } fclose(fp1); printf("\n\tType the text and press 'Ctrl+s' to append.\n"); fp1=fopen(fn,"a"); while(1) { c=getchar(); if(c==19) goto end3; if(c==13) { d='\n'; fputc(d,fp1); } else { fputc(c,fp1); } } end3: fclose(fp1); }

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• MySql multiple selects batching in .net

  - by Amith George

  I have a situation in my application. For each x-axis point in my chart, I am plotting 5 y-axis values. To calculate each of these 5 values, I need to make 4 different queries. Ie, for each x-axis point I need to fire 20 sql queries. Now, I need to plot 40 such points in the my chart. Its resulting in a pathetic performance where it takes close to a minute to get all the data back from the database. Each of 4 different queries consists of a join between 2 tables. One has only 6 rows. The other close to 10,000. Each of the 4 queries has different WHERE clauses, so they are different queries. For each point in the x-axis, only the values for the where clauses change. I have tried combining each of the 4 queries into one big string. Basically batch the four selects. These are again batched for each y-axis value. So, for each x-axis point, I am now firing one big command that consists of 20 different select statements. Technically, I should be experiencing a big performance boost, right? Instead of hitting the db 40x5x4 = 800 times, I am now hitting it just 40 times. But instead of taking 60 seconds, it taking 50-55 seconds... not much of a help. I am using MySql 5.1, and the 6.1 version of its .Net connector. What can I do to improve the performance? Edit: One of the 4 queries is as follows: SELECT SUM(TIME_TO_SEC(TIMEDIFF(T1.col2, T1.col1))* T2.col1 / (3600 *1000)) AS TotalTime FROM Table T1 JOIN Table T2 ON T1.col3 = T2.col3 WHERE T1.col4 = 'i' AND T1.col1 >= '2009-12-25 00:00:00' AND T1.col2 <= '2009-12-26 00:00:00'; The other 3 queries are similar, only the where clause changes slightly. This set of 4 queries is fired 5 times. The first 3 times against the join of table T1 and T2, passing in different values for col4. And the next two times against the join of table T3 and T2 passing in different values for col4. These 5 values are the y-axis values for a particular x-axis point. The data returned by all these queries is the same format. so, we tried doing a UNION ALL on all these queries. No substantial difference. One strange thing, however, after indexing the foreign key on the table T1 [while it contained over a lakh records], the queries were using the index, but they had become slower. At times, the queries would take double the time to return the data.

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• EJB client can't find a DataSource that tests successfully in WebLogic admin console

  - by suszterpatt

  Disclaimer: I'm completely new to JEE/EJB and all that, so bear with me. I have a simple EJB that I can successfully deploy using JDeveloper 11g's integrated WebLogic server and a remote database connection (JDBC). I have a DataSource named "PGY2" defined in WebLogic, and I can test it successfully from the admin console. Here's the code of the client I'm trying to test it with (generated entirely by JDev except for the three method calls on adminManager): public class AdminManagerClient { public static void main(String [] args) { try { final Context context = getInitialContext(); AdminManager adminManager = (AdminManager)context.lookup("Uran-AdminManager#hu.elte.pgy2.BACNAAI.UranEJB.AdminManager"); adminManager.addAdmin("root", "root", "Kovács Isten"); adminManager.addStudent("BACNAAI", "matt", "B Cs", 2005); adminManager.addTeacher("SIPKABT", "patt", "S P", "numanal", "Dr."); } catch (Exception ex) { ex.printStackTrace(); } } private static Context getInitialContext() throws NamingException { Hashtable env = new Hashtable(); // WebLogic Server 10.x connection details env.put( Context.INITIAL_CONTEXT_FACTORY, "weblogic.jndi.WLInitialContextFactory" ); env.put(Context.PROVIDER_URL, "t3://127.0.0.1:7101"); return new InitialContext( env ); } } But when I try to run this, I get the following error on the line with adminManager.addAdmin (that is, after the lookup): javax.ejb.EJBException: EJB Exception: ; nested exception is: Exception [EclipseLink-4002] (Eclipse Persistence Services - 1.0.2 (Build 20081024)): org.eclipse.persistence.exceptions.DatabaseException Internal Exception: java.sql.SQLException: Internal error: Cannot obtain XAConnection Creation of XAConnection for pool PGY2 failed after waitSecs:30 : java.sql.SQLException: Data Source PGY2 does not exist. Why can't the client find the data source, and how do I make it find it? EDIT: I took a closer look at WebLogic's output during deployment, and I found this. I have no idea what it means, but it may be relevant: <2010.05.20. 0:50:43 CEST> <Error> <Deployer> <BEA-149231> <Unable to set the activation state to true for the application 'PGY2'. weblogic.application.ModuleException: at weblogic.jdbc.module.JDBCModule.activate(JDBCModule.java:349) at weblogic.application.internal.flow.ModuleListenerInvoker.activate(ModuleListenerInvoker.java:107) at weblogic.application.internal.flow.DeploymentCallbackFlow$2.next(DeploymentCallbackFlow.java:411) at weblogic.application.utils.StateMachineDriver.nextState(StateMachineDriver.java:37) at weblogic.application.internal.flow.DeploymentCallbackFlow.activate(DeploymentCallbackFlow.java:74) Truncated. see log file for complete stacktrace weblogic.common.ResourceException: is already bound at weblogic.jdbc.common.internal.RmiDataSource.start(RmiDataSource.java:387) at weblogic.jdbc.common.internal.DataSourceManager.createAndStartDataSource(DataSourceManager.java:136) at weblogic.jdbc.common.internal.DataSourceManager.createAndStartDataSource(DataSourceManager.java:97) at weblogic.jdbc.module.JDBCModule.activate(JDBCModule.java:346) at weblogic.application.internal.flow.ModuleListenerInvoker.activate(ModuleListenerInvoker.java:107) Truncated. see log file for complete stacktrace >

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• Oracle Schema Design: Seperate Schema with I/O Overhead?

  - by Guru

  We are designing database schema for a new system based on Oracle 11gR1. We have identified a main schema which would have close to 100 tables, these will be accessed from the front end Java application. We have a requirement to audit the values which got changed in close to 50 tables, this has to be done every row. Which means, it is possible that, for a single row in MYSYS.T1 there might be 50 (or more) rows in MYSYS_AUDIT.T1_AUD table. We might be having old values of every column entry and new values available from T1. DBA gave an observation, advising against this method, because he said, separate schema meant an extra I/O for every operation. Basically AUDIT schema would be used only to do some analyse and enter values (thus SELECT and INSERT). Is it true that, "a separate schema means an extra I/O" ? I could not find justification. It appears logical to me, as the AUDIT data should not be tampered with, so a separate schema. Also, we designed a separate schema for archiving some tables from MYSYS. From MYSYS_ARC the table might be backed up into tapes or deleted after sufficient time. Few stats: Few tables (close to 20, 30) in MYSYS schema could grow to around 50M rows. We have asked for a total disk space of 4 TB. MYSYS_AUDIT schema might be having 10 times that of MYSYS but we wont keep them more than 3 months. Questions Given all these, can you suggest me any improvements? Separate schema affects disc I/O? (one extra I/O for every schema ?) Any general suggestions? Figure: +-------------------+ +-------------------+ | MYSYS | | MYSYS_AUDIT | | | | | | 1. T1 | | 1. T1_AUD | | 2. T2 | | 2. T2_AUD | | 3. T3 |--------->| 3. T3_AUD | | 4. T4 |(SELECT, | 4. T4_AUD | | . | INSERT) | . | | . | | . | | . | | . | | 100. T100 | | 50. T50_AUD | +-------------------+ +-------------------+ | | | | |(INSERT) | | | * +-------------------+ | MYSYS_ARC | | | | 1. T1_ARC | | 2. T2_ARC | | 3. T3_ARC | | 4. T4_ARC | | . | | . | | . | | 100. T100_ARC | +-------------------+ Apart from this, we have two more schemas with only read only rights, but mainly they are for adhoc purpose and we dont mind the performance on them.

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• how to find maximum frequent item sets from large transactional data file

  - by ANIL MANE

  Hi, I have the input file contains large amount of transactions like Transaction ID Items T1 Bread, milk, coffee, juice T2 Juice, milk, coffee T3 Bread, juice T4 Coffee, milk T5 Bread, Milk T6 Coffee, Bread T7 Coffee, Bread, Juice T8 Bread, Milk, Juice T9 Milk, Bread, Coffee, T10 Bread T11 Milk T12 Milk, Coffee, Bread, Juice i want the occurrence of every unique item like Item Name Count Bread 9 Milk 8 Coffee 7 Juice 6 and from that i want an a fp-tree now by traversing this tree i want the maximal frequent itemsets as follows The basic idea of method is to dispose nodes in each “layer” from bottom to up. The concept of “layer” is different to the common concept of layer in a tree. Nodes in a “layer” mean the nodes correspond to the same item and be in a linked list from the “Head Table”. For nodes in a “layer” NBN method will be used to dispose the nodes from left to right along the linked list. To use NBN method, two extra fields will be added to each node in the ordered FP-Tree. The field tag of node N stores the information of whether N is maximal frequent itemset, and the field count’ stores the support count information in the nodes at left. In Figure, the first node to be disposed is “juice: 2”. If the min_sup is equal to or less than 2 then “bread, milk, coffee, juice” is a maximal frequent itemset. Firstly output juice:2 and set the field tag of “coffee:3” as “false” (the field tag of each node is “true” initially ). Next check whether the right four itemsets juice:1 be the subset of juice:2. If the itemset one node “juice:1” corresponding to is the subset of juice:2 set the field tag of the node “false”. In the following process when the field tag of the disposed node is FALSE we can omit the node after the same tagging. If the min_sup is more than 2 then check whether the right four juice:1 is the subset of juice:2. If the itemset one node “juice:1” corresponding to is the subset of juice:2 then set the field count’ of the node with the sum of the former count’ and 2 After all the nodes “juice” disposed ,begin to dispose the node “coffee:3”. Any suggestions or available source code, welcome. thanks in advance

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• chaining array of tasks with continuation

  - by Andrei Cristof

  I have a Task structure that is a little bit complex(for me at least). The structure is: (where T = Task) T1, T2, T3... Tn. There's an array (a list of files), and the T's represent tasks created for each file. Each T has always two subtasks that it must complete or fail: Tn.1, Tn.2 - download and install. For each download (Tn.1) there are always two subtasks to try, download from two paths(Tn.1.1, Tn.1.2). Execution would be: First, download file: Tn1.1. If Tn.1.1 fails, then Tn.1.2 executes. If either from download tasks returns OK - execute Tn.2. If Tn.2 executed or failed - go to next Tn. I figured the first thing to do, was to write all this structure with jagged arrays: private void CreateTasks() { //main array Task<int>[][][] mainTask = new Task<int>[_queuedApps.Count][][]; for (int i = 0; i < mainTask.Length; i++) { Task<int>[][] arr = GenerateOperationTasks(); mainTask[i] = arr; } } private Task<int>[][] GenerateOperationTasks() { //two download tasks Task<int>[] downloadTasks = new Task<int>[2]; downloadTasks[0] = new Task<int>(() => { return 0; }); downloadTasks[1] = new Task<int>(() => { return 0; }); //one installation task Task<int>[] installTask = new Task<int>[1] { new Task<int>(() => { return 0; }) }; //operations Task is jagged - keeps tasks above Task<int>[][] operationTasks = new Task<int>[2][]; operationTasks[0] = downloadTasks; operationTasks[1] = installTask; return operationTasks; } So now I got my mainTask array of tasks, containing nicely ordered tasks just as described above. However after reading the docs on ContinuationTasks, I realise this does not help me since I must call e.g. Task.ContinueWith(Task2). I'm stumped about doing this on my mainTask array. I can't write mainTask[0].ContinueWith(mainTask[1]) because I dont know the size of the array. If I could somehow reference the next task in the array (but without knowing its index), but cant figure out how. Any ideas? Thank you very much for your help. Regards,

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• How to use constraint programming for optimizing shopping baskets?

  - by tangens

  I have a list of items I want to buy. The items are offered by different shops and different prices. The shops have individual delivery costs. I'm looking for an optimal shopping strategy (and a java library supporting it) to purchase all of the items with a minimal total price. Example: Item1 is offered at Shop1 for $100, at Shop2 for $111. Item2 is offered at Shop1 for $90, at Shop2 for $85. Delivery cost of Shop1: $10 if total order < $150; $0 otherwise Delivery cost of Shop2: $5 if total order < $50; $0 otherwise If I buy Item1 and Item2 at Shop1 the total cost is $100 + $90 +$0 = $190. If I buy Item1 and Item2 at Shop2 the total cost is $111 + $85 +$0 = $196. If I buy Item1 at Shop1 and Item2 at Shop2 the total cost is $100 + $10 + $85 + $0 = 195. I get the minimal price if I order Item1 and Item2 at Shop1: $190 What I tried so far I asked another question before that led me to the field of constraint programming. I had a look at cream and choco, but I did not figure out how to create a model to solve my problem. | shop1 | shop2 | shop3 | ... ----------------------------------------- item1 | p11 | p12 | p13 | item2 | p21 | p22 | p23 | . | | | | . | | | | ----------------------------------------- shipping | s1 | s2 | s3 | limit | l1 | l2 | l3 | ----------------------------------------- total | t1 | t2 | t3 | ----------------------------------------- My idea was to define these constraints: each price "p xy" is defined in the domain (0, c) where c is the price of the item in this shop only one price in a line should be non zero if one or more items are bought from one shop and the sum of the prices is lower than limit, then add shipping cost to the total cost shop total cost is the sum of the prices of all items in a shop total cost is the sum of all shop totals The objective is "total cost". I want to minimize this. In cream I wasn't able to express the "if then" constraint for conditional shipping costs. In choco these constraints exist, but even for 5 items and 10 shops the program was running for 10 minutes without finding a solution. Question How should I express my constraints to make this problem solvable for a constraint programming solver?

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