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• What is the effect of this order_by clause?

  - by bread

  I don't understand what this order_by clause is doing and whether I need it or not: select c.customerid, c.firstname, c.lastname, i.order_date, i.item, i.price from items_ordered i, customers c where i.customerid = c.customerid group by c.customerid, i.item, i.order_date order by i.order_date desc; This produces this data: 10330 Shawn Dalton 30-Jun-1999 Pogo stick 28.00 10101 John Gray 30-Jun-1999 Raft 58.00 10410 Mary Ann Howell 30-Jan-2000 Unicycle 192.50 10101 John Gray 30-Dec-1999 Hoola Hoop 14.75 10449 Isabela Moore 29-Feb-2000 Flashlight 4.50 10410 Mary Ann Howell 28-Oct-1999 Sleeping Bag 89.22 10339 Anthony Sanchez 27-Jul-1999 Umbrella 4.50 10449 Isabela Moore 22-Dec-1999 Canoe 280.00 10298 Leroy Brown 19-Sep-1999 Lantern 29.00 10449 Isabela Moore 19-Mar-2000 Canoe paddle 40.00 10413 Donald Davids 19-Jan-2000 Lawnchair 32.00 10330 Shawn Dalton 19-Apr-2000 Shovel 16.75 10439 Conrad Giles 18-Sep-1999 Tent 88.00 10298 Leroy Brown 18-Mar-2000 Pocket Knife 22.38 10299 Elroy Keller 18-Jan-2000 Inflatable Mattress 38.00 10438 Kevin Smith 18-Jan-2000 Tent 79.99 10101 John Gray 18-Aug-1999 Rain Coat 18.30 10449 Isabela Moore 15-Dec-1999 Bicycle 380.50 10439 Conrad Giles 14-Aug-1999 Ski Poles 25.50 10449 Isabela Moore 13-Aug-1999 Unicycle 180.79 10101 John Gray 08-Mar-2000 Sleeping Bag 88.70 10299 Elroy Keller 06-Jul-1999 Parachute 1250.00 10438 Kevin Smith 02-Nov-1999 Pillow 8.50 10101 John Gray 02-Jan-2000 Lantern 16.00 10315 Lisa Jones 02-Feb-2000 Compass 8.00 10449 Isabela Moore 01-Sep-1999 Snow Shoes 45.00 10438 Kevin Smith 01-Nov-1999 Umbrella 6.75 10298 Leroy Brown 01-Jul-1999 Skateboard 33.00 10101 John Gray 01-Jul-1999 Life Vest 125.00 10330 Shawn Dalton 01-Jan-2000 Flashlight 28.00 10298 Leroy Brown 01-Dec-1999 Helmet 22.00 10298 Leroy Brown 01-Apr-2000 Ear Muffs 12.50 While if I remove the order_by clause completely, as in this query: select c.customerid, c.firstname, c.lastname, i.order_date, i.item, i.price from items_ordered i, customers c where i.customerid = c.customerid group by c.customerid, i.item, i.order_date; I get these results: 10101 John Gray 30-Dec-1999 Hoola Hoop 14.75 10101 John Gray 02-Jan-2000 Lantern 16.00 10101 John Gray 01-Jul-1999 Life Vest 125.00 10101 John Gray 30-Jun-1999 Raft 58.00 10101 John Gray 18-Aug-1999 Rain Coat 18.30 10101 John Gray 08-Mar-2000 Sleeping Bag 88.70 10298 Leroy Brown 01-Apr-2000 Ear Muffs 12.50 10298 Leroy Brown 01-Dec-1999 Helmet 22.00 10298 Leroy Brown 19-Sep-1999 Lantern 29.00 10298 Leroy Brown 18-Mar-2000 Pocket Knife 22.38 10298 Leroy Brown 01-Jul-1999 Skateboard 33.00 10299 Elroy Keller 18-Jan-2000 Inflatable Mattress 38.00 10299 Elroy Keller 06-Jul-1999 Parachute 1250.00 10315 Lisa Jones 02-Feb-2000 Compass 8.00 10330 Shawn Dalton 01-Jan-2000 Flashlight 28.00 10330 Shawn Dalton 30-Jun-1999 Pogo stick 28.00 10330 Shawn Dalton 19-Apr-2000 Shovel 16.75 10339 Anthony Sanchez 27-Jul-1999 Umbrella 4.50 10410 Mary Ann Howell 28-Oct-1999 Sleeping Bag 89.22 10410 Mary Ann Howell 30-Jan-2000 Unicycle 192.50 10413 Donald Davids 19-Jan-2000 Lawnchair 32.00 10438 Kevin Smith 02-Nov-1999 Pillow 8.50 10438 Kevin Smith 18-Jan-2000 Tent 79.99 10438 Kevin Smith 01-Nov-1999 Umbrella 6.75 10439 Conrad Giles 14-Aug-1999 Ski Poles 25.50 10439 Conrad Giles 18-Sep-1999 Tent 88.00 10449 Isabela Moore 15-Dec-1999 Bicycle 380.50 10449 Isabela Moore 22-Dec-1999 Canoe 280.00 10449 Isabela Moore 19-Mar-2000 Canoe paddle 40.00 10449 Isabela Moore 29-Feb-2000 Flashlight 4.50 10449 Isabela Moore 01-Sep-1999 Snow Shoes 45.00 10449 Isabela Moore 13-Aug-1999 Unicycle 180.79 I'm not sure what the order_by is doing here and if it's having the intended effects.

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• Tricky SQL query involving consecutive values

  - by Gabriel

  I need to perform a relatively easy to explain but (given my somewhat limited skills) hard to write SQL query. Assume we have a table similar to this one: exam_no | name | surname | result | date ---------+------+---------+--------+------------ 1 | John | Doe | PASS | 2012-01-01 1 | Ryan | Smith | FAIL | 2012-01-02 <-- 1 | Ann | Evans | PASS | 2012-01-03 1 | Mary | Lee | FAIL | 2012-01-04 ... | ... | ... | ... | ... 2 | John | Doe | FAIL | 2012-02-01 <-- 2 | Ryan | Smith | FAIL | 2012-02-02 2 | Ann | Evans | FAIL | 2012-02-03 2 | Mary | Lee | PASS | 2012-02-04 ... | ... | ... | ... | ... 3 | John | Doe | FAIL | 2012-03-01 3 | Ryan | Smith | FAIL | 2012-03-02 3 | Ann | Evans | PASS | 2012-03-03 3 | Mary | Lee | FAIL | 2012-03-04 <-- Note that exam_no and date aren't necessarily related as one might expect from the kind of example I chose. Now, the query that I need to do is as follows: From the latest exam (exam_no = 3) find all the students that have failed (John Doe, Ryan Smith and Mary Lee). For each of these students find the date of the first of the batch of consecutively failing exams. Another way to put it would be: for each of these students find the date of the first failing exam that comes after their last passing exam. (Look at the arrows in the table). The resulting table should be something like this: name | surname | date_since_failing ------+---------+-------------------- John | Doe | 2012-02-01 Ryan | Smith | 2012-01-02 Mary | Lee | 2012-01-04 Ann | Evans | 2012-02-03 How can I perform such a query? Thank you for your time.

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• Odd SQL Results

  - by Ryan Burnham

  So i have the following query Select id, [First], [Last] , [Business] as contactbusiness, (Case When ([Business] != '' or [Business] is not null) Then [Business] Else 'No Phone Number' END) from contacts The results look like id First Last contactbusiness (No column name) 2 John Smith 3 Sarah Jane 0411 111 222 0411 111 222 6 John Smith 0411 111 111 0411 111 111 8 NULL No Phone Number 11 Ryan B 08 9999 9999 08 9999 9999 14 David F NULL No Phone Number I'd expect record 2 to also show No Phone Number If i change the "[Business] is not null" to [Business] != null then i get the correct results id First Last contactbusiness (No column name) 2 John Smith No Phone Number 3 Sarah Jane 0411 111 222 0411 111 222 6 John Smith 0411 111 111 0411 111 111 8 NULL No Phone Number 11 Ryan B 08 9999 9999 08 9999 9999 14 David F NULL No Phone Number Normally you need to use is not null rather than != null. whats going on here?

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• Creating an excel macro to sum lines with duplicate values

  - by john

  I need a macro to look at the list of data below, provide a number of instances it appears and sum the value of each of them. I know a pivot table or series of forumlas could work but i'm doing this for a coworker and it has to be a 'one click here' kinda deal. The data is as follows. A B Smith 200.00 Dean 100.00 Smith 100.00 Smith 50.00 Wilson 25.00 Dean 25.00 Barry 100.00 The end result would look like this Smith 3 350.00 Dean 2 125.00 Wilson 1 25.00 Barry 1 100.00 Thanks in advance for any help you can offer!

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• Simple Select Statement on MySQL Database Hanging

  - by AlishahNovin

  I have a very simple sql select statement on a very large table, that is non-normalized. (Not my design at all, I'm just trying to optimize while simultaneously trying to convince the owners of a redesign) Basically, the statement is like this: SELECT FirstName, LastName, FullName, State FROM Activity Where (FirstName=@name OR LastName=@name OR FullName=@name) AND State=@state; Now, FirstName, LastName, FullName and State are all indexed as BTrees, but without prefix - the whole column is indexed. State column is a 2 letter state code. What I'm finding is this: When @name = 'John Smith', and @state = '%' the search is really fast and yields results immediately. When @name = 'John Smith', and @state = 'FL' the search takes 5 minutes (and usually this means the web service times out...) When I remove the FirstName and LastName comparisons, and only use the FullName and State, both cases above work very quickly. When I replace FirstName, LastName, FullName, and State searches, but use LIKE for each search, it works fast for @name='John Smith%' and @state='%', but slow for @name='John Smith%' and @state='FL' When I search against 'John Sm%' and @state='FL' the search finds results immediately When I search against 'John Smi%' and @state='FL' the search takes 5 minutes. Now, just to reiterate - the table is not normalized. The John Smith appears many many times, as do many other users, because there is no reference to some form of users/people table. I'm not sure how many times a single user may appear, but the table itself has 90 Million records. Again, not my design... What I'm wondering is - though there are many many problems with this design, what is causing this specific problem. My guess is that the index trees are just too large that it just takes a very long time traversing the them. (FirstName, LastName, FullName) Anyway, I appreciate anyone's help with this. Like I said, I'm working on convincing them of a redesign, but in the meantime, if I someone could help me figure out what the exact problem is, that'd be fantastic.

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• Is it possible to use ContainsTable to get results for more than one column?

  - by LockeCJ

  Consider the following table: People FirstName nvarchar(50) LastName nvarchar(50) Let's assume for the moment that this table has a full-text index on it for both columns. Let's suppose that I wanted to find all of the people named "John Smith" in this table. The following query seems like a perfectly rational way to accomplish this: SELECT * from People p INNER JOIN CONTAINSTABLE(People,*,'"John*" AND "Smith*"') Unfortunately, this will return no results, assuming that there is no record in the People table that contains both "John" and "Smith" in either the FirstName or LastName columns. It will not match a record with "John" in the FirstName column, and "Smith" in the LastName column, or vice-versa. My question is this: How does one accomplish what I'm trying to do above? Please consider that the example above is simplified. The real table I'm working with has ten columns and the input I'm receiving is a single string which is split up based on standard word breakers (space, dash, etc.)

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• Parsing complex string using regex

  - by wojtek_z

  My regex skills are not very good and recently a new data element has thrown my parser into a loop Take the following string "+USER=Bob Smith-GROUP=Admin+FUNCTION=Read/FUNCTION=Write" Previously I had the following for my regex : [+\\-/] Which would turn the result into USER=Bob Smith GROUP=Admin FUNCTION=Read FUNCTION=Write FUNCTION=Read But now I have values with dashes in them which is causing bad output New string looks like "+USER=Bob Smith-GROUP=Admin+FUNCTION=Read/FUNCTION=Write/FUNCTION=Read-Write" Which gives me the following result , and breaks the key = value structure. USER=Bob Smith GROUP=Admin FUNCTION=Read FUNCTION=Write FUNCTION=Read Write Can someone help me formulate a valid regex for handling this or point me to some key / value examples. Basically I need to be able to handle + - / signs in order to get combinations.

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• Are there any well known algorithms to detect the presence of names?

  - by Rhubarb

  For example, given a string: "Bob went fishing with his friend Jim Smith." Bob and Jim Smith are both names, but bob and smith are both words. Weren't for them being uppercase, there would be less indication of this outside of our knowledge of the sentence. Without doing grammar analysis, are there any well known algorithms for detecting the presence of names, at least Western names?

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• Need help joining tables...

  - by yuudachi

  I am a MySQL newbie, so sorry if this is a dumb question.. These are my tables. student table: SID (primary) student_name advisor (foreign key to faculty.facultyID) requested_advisor (foreign key to faculty.facultyID) faculty table: facultyID (primary key) advisor_name I want to query a table that shows everything in the student table, but I want advisor and requested_advisor to show up as names, not the ID numbers. so like it displays like this on the webpage: Student Name: Jane Smith SID: 860123456 Current Advisor: John Smith Requested advisor: James Smith not like this Student Name: Jane Smith SID: 860123456 Current Advisor: 1 Requested advisor: 2 SELECT student.student_name, SID, student_email, faculty.advisor_name FROM student INNER JOIN faculty ON student.advisor = faculty.facultyID; this comes out close, but I don't know how to get the requested_advisor to show up as a name.

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• Setting variables in shell script by running commands

  - by rajya vardhan

  >cat /tmp/list1 john jack >cat /tmp/list2 smith taylor It is guaranteed that list1 and list2 will have equal number of lines. f(){ i=1 while read line do var1 = `sed -n '$ip' /tmp/list1` var2 = `sed -n '$ip' /tmp/list2` echo $i,$var1,$var2 i=`expr $i+1` echo $i,$var1,$var2 done < $INFILE } So output of f() should be: 1,john,smith 2,jack,taylor But getting 1,p,p 1+1,p,p If i replace following: var1 = `sed -n '$ip' /tmp/list1` var2 = `sed -n '$ip' /tmp/list2` with this: var1=`head -$i /tmp/vip_list|tail -1` var2=`head -$i /tmp/lb_list|tail -1` Then output: 1,john,smith 1,john,smith Not an expert of shell, so please excuse if sounds childish :)

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• Creating one row of information in excel using a unique value

  - by user1426513

  This is my first post. I am currently working on a project at work which requires that I work with several different worksheets in order to create one mail master worksheet, as it were, in order to do a mail merge. The worksheet contains information regarding different purchases, and each purchaser is identified with their own ID number. Below is an example of what my spreadsheet looks like now (however I do have more columns): ID Salutation Address ID Name Donation ID Name Tickets 9 Mr. John Doe 123 12 Ms. Jane Smith 100.00 12 Ms.Jane Smith 300.00 12 Ms. Jane Smith 456 22 Mr. Mike Man 500.00 84 Ms. Jo Smith 300.00 What I would like to do is somehow sort my data so that everythign with the same unique identifier (ID) lines up on the same row. For example ID 12 Jane Smith - all the information for her will show up under her name matched by her ID number, and ID 22 will match up with 22 etc... When I merged all of my spreadsheets together, I sorted them all by ID number, however my problem is, not everyone who made a donation bought a ticket or some people just bought tickets and nothing us, so sorting doesn't work. Hopefully this makes sense. Thanks in advance.

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• Run script after switching user account "to the same account"

  - by Peter Sivák

  In Ubuntu, when I click on Switch User Account... and then choose the same account to log in (for example if my name is John Smith, I click on switch user account and then log into the John Smith account again), how can I run a script after that? (I know, that I can run a script after "first" login by putting it in /etc/profile file, but this script is not executed again when I choose switch user account and then immediately log in back to the same account.)

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• GDL Presents: Women Techmakers with JESS3

  GDL Presents: Women Techmakers with JESS3 Join Leslie, COO and Co-founder of JESS3, in conversation with Megan Smith and Betsy Masiello, as they discuss Leslie's experience growing a design business from two employees to a transnational operation. Hosts: Megan Smith - Vice President, Google [x] | Betsy Masiello - Policy Manager Guest: Leslie Bradshaw - President, COO and Co-founder, JESS3 From: GoogleDevelopers Views: 0 3 ratings Time: 01:00:00 More in Science & Technology

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• How to count the most recent value based on multiple criteria?

  - by Andrew

  I keep a log of phone calls like the following where the F column is LVM = Left Voice Mail, U = Unsuccessful, S = Successful. A1 1 B1 Smith C1 John D1 11/21/2012 E1 8:00 AM F1 LVM A2 2 B2 Smith C2 John D2 11/22/2012 E1 8:15 AM F2 U A3 3 B3 Harvey C3 Luke D3 11/22/2012 E1 8:30 AM F3 S A4 4 B4 Smith C4 John D4 11/22/2012 E1 9:00 AM F4 S A5 5 B5 Smith C5 John D5 11/23/2012 E5 8:00 AM F5 LVM This is a small sample. I actually have over 700 entries. In my line of work, it is important to know how many unsuccessful (LVM or U) calls I have made since the last Successful one (S). Since values in the F column can repeat, I need to take into consideration both the B and C column. Also, since I can make a successful call with a client and then be trying to contact them again, I need to be able to count from the last successful call. My G column is completely open which is where I would like to put a running total for each client (G5 would = 1 ideally while G4 = 0, G3 = 0, G2 = 2, G1 = 1 but I want these values calculated automatically so that I do not have scroll through 700 names).

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• MS Word 2010: Hide citation title when 2 publications by same first author from different years are in one citation block

  - by srunni

  I'm trying to hide the display of the titles for two publications by the same first author from different years that are in the same citation block. By default, the title is shown in citations when there are two publications by the same author in a given document. The easiest way to get around this is to right click on the citation, click "Edit Citation", and then suppress the title. However, the issue with this is that if there are 2 citations in 1 citation block (i.e., "(Smith, J., et al. 2010, Smith, J., et al. 2011)" rather than "(Smith, J., et al. 2010) (Smith, J., et al. 2011)"), then using that suppress option only suppresses the title for the first citation (in this case, the 2010 publication). OTOH, if I try to initially insert the publications in separate citation blocks, I can suppress the title in both citations, but I can't cut and paste one into the other's citation block. I can click "Cut" and the citation that was just cut disappears, but the "Paste" option is not available when my cursor is in the second citation block. Any ideas? Thanks!

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• To ref or not to ref

  - by nmarun

  So the question is what is the point of passing a reference type along with the ref keyword? I have an Employee class as below: 1: public class Employee 2: { 3: public string FirstName { get; set; } 4: public string LastName { get; set; } 5:  6: public override string ToString() 7: { 8: return string.Format("{0}-{1}", FirstName, LastName); 9: } 10: } In my calling class, I say: 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(Employee employee) 16: { 17: employee.FirstName = "Smith"; 18: employee.LastName = "Doe"; 19: } 20: }   After having a look at the code, you’ll probably say, Well, an instance of a class gets passed as a reference, so any changes to the instance inside the CallSomeMethod, actually modifies the original object. Hence the output will be ‘John-Doe’ on the first call and ‘Smith-Doe’ on the second. And you’re right: So the question is what’s the use of passing this Employee parameter as a ref? 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(ref employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(ref Employee employee) 16: { 17: employee.FirstName = "Smith"; 18: employee.LastName = "Doe"; 19: } 20: } The output is still the same: Ok, so is there really a need to pass a reference type using the ref keyword? I’ll remove the ‘ref’ keyword and make one more change to the CallSomeMethod method. 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(Employee employee) 16: { 17: employee = new Employee 18: { 19: FirstName = "Smith", 20: LastName = "John" 21: }; 22: } 23: } In line 17 you’ll see I’ve ‘new’d up the incoming Employee parameter and then set its properties to new values. The output tells me that the original instance of the Employee class does not change. Huh? But an instance of a class gets passed by reference, so why did the values not change on the original instance or how do I keep the two instances in-sync all the times? Aah, now here’s the answer. In order to keep the objects in sync, you pass them using the ‘ref’ keyword. 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(ref employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(ref Employee employee) 16: { 17: employee = new Employee 18: { 19: FirstName = "Smith", 20: LastName = "John" 21: }; 22: } 23: } Viola! Now, to prove it beyond doubt, I said, let me try with another reference type: string. 1: class Program 2: { 3: static void Main() 4: { 5: string name = "abc"; 6: Console.WriteLine(name); 7: CallSomeMethod(ref name); 8: Console.WriteLine(name); 9: } 10:  11: private static void CallSomeMethod(ref string name) 12: { 13: name = "def"; 14: } 15: } The output was as expected, first ‘abc’ and then ‘def’ - proves the 'ref' keyword works here as well. Now, what if I remove the ‘ref’ keyword? The output should still be the same as the above right, since string is a reference type? 1: class Program 2: { 3: static void Main() 4: { 5: string name = "abc"; 6: Console.WriteLine(name); 7: CallSomeMethod(name); 8: Console.WriteLine(name); 9: } 10:  11: private static void CallSomeMethod(string name) 12: { 13: name = "def"; 14: } 15: } Wrong, the output shows ‘abc’ printed twice. Wait a minute… now how could this be? This is because string is an immutable type. This means that any time you modify an instance of string, new memory address is allocated to the instance. The effect is similar to ‘new’ing up the Employee instance inside the CallSomeMethod in the absence of the ‘ref’ keyword. Verdict: ref key came to the rescue and saved the planet… again!

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• Change XRDP keyboard layout to en-gb Ubuntu 12.04

  - by Earl Sven

  Does anybody know how to change the keyboard layout to en-gb in an XRDP session on Ubuntu 12.04? I am using mstsc.exe to connect to an XRDP server hosting an XVNC session, however I cannot work out how to apply the UK keyboard layout. A bit of googling has yeilded these instructions which allow me to change the keymap, however using the keymap file I downloaded from here I loose the ability to use the arrow keys, home/end etc. Comparing the file with the standard one there are substantially more differences than I would expect considering the similarity between the layouts. I only have RDP access to the box so i don't seem to be able to actually generate a new layout per the instructions above, maybe it's a local console thing? Also I can't change either the RDP client used or the RDP server as they are my only access to the system, I don't have local console access. I do have root priveleges on the OS however. Any thoughts? Edit: I have found http:// xrdp.sourceforge.net/documents/keymap/newkeymap.html (apologies for not typing the link properly but the antispam filter won't let me post more than 2 links) this documentation on the XRDP sourceforge page which describes keymap file format. It indicates the values in the keymap files are unicode 0x64 etc, however the files I have already on my system seem to use a different format 0:0 or 65307:27 etc, does anybody know what the difference is?

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• Server have 2 psu, can i only turn on 1 psu, to reduce cost in colocation?

  - by Earl

  i just got a server & want to colocation it in datacenter server details : HP DL380, 2x intel Xeon (3,06GHz/533, 512KB L2 Cache), 8x Fans, Form Factor Rack (2U), 2x 400W Power Supplies, the server have 2 psu, can i only turn on 1 psu, to reduce cost in colocation? will the server still running good? the standart colocation packages in my city only give default power 400w, if need additional power 400w need additional cost about $40-60 again permonth please give suggestion from your experience

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• Adobe CS5.1; I want the measure tool as the default; when I use Ctrl-I, the measure (ruler) tool appears... is this sort of thing possible?

  - by Earl J

  I scan many slides, transparencies, and negatives. I need to straighten them often. I would like the ruler/measurement tool under CTRL-I to remain as the default with eveery start of Photoshop. Is it possible to make it the default instead of the eyedropper tool? Perhaps a shortcut key of its own might work as well... where one keystroke will bring it to the top... will it stay there? hmmm. . . I've just given myself an idea... (LOL)

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• It's Not TV- It's OTN: Top 10 Videos on the OTN YouTube Channel

  - by Bob Rhubart

  It's been a while since we checked in on what people are watching on the Oracle Technology Network YouTube Channel. Here are the Top 10 video for the last 30 days. Tom Kyte: Keeping Up with the Latest in Database Technology Tom Kyte expands on his keynote presentation at the Great Lakes Oracle Conference with tips for developers, DBAs and others who want to make sure they are prepared to work with the latest database technologies. That Jeff Smith: Oracle SQL Developer Oracle SQL Developer product manager Jeff Smith (yeah, that Jeff Smith) talks about his presentations at the Great Lakes Oracle Conference and shares his reaction to keynote speaker C.J. Date's claim that "SQL dropped the ball." Gwen Shapira: Hadoop and Oracle Database Oracle ACE Director Gwen Shapira @gwenshap talks about the fit between Hadoop and Oracle Database and dives into the details of why Oracle Loader for Hadoop is 5x faster. Kai Yu: Virtualization and Cloud Oracle ACE Director Kai Yu talks about the questions he is most frequently asked when he does presentations on cloud computing and virtualization. Mark Sewtz: APEX 4.2 Mobile App Development Application Express developer Marc Sewtz demos the new features he built into APEX4.2 to support Mobile App Development. Jeremy Schneider: RAC Attack Oracle ACE Jeremy Schneider @jer_s describes what you can expect when you come to a RAC (Real Application Cluster) Attack. Frits Hoogland: Exadata Under the Hood Oracle ACE Director Frits Hoogland (@fritshoogland) talks about the secret sauce under Exadata's hood. David Peake: APEX 4.2 New Features David Peake, PM for Oracle Application Express, gives a quick overview of some of the new APEX features. Greg Marsden: Hugepages = Huge Performance on Linux Greg Marsden of Oracle's Linux Kernel Engineering Team talks about some common customer performance questions and making the most of Oracle Linux 6 and Transparent HugePages. John Hurley: NEOOUG and GLOC 2013 Northeast Ohio Oracle User Group president John Hurley talks about the background and success of the 2013 Great Lakes Oracle Conference.

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• Password Policy seems to be ignored for new Domain on Windows Server 2008 R2

  - by Earl Sven

  I have set up a new Windows Server 2008 R2 domain controller, and have attempted to configure the Default Domain Policy to permit all types of passwords. When I want to create a new user (just a normal user) in the Domain Users and Computers application, I am prevented from doing so because of password complexity/length reasons. The password policy options configured in the Default Domain Policy are not defined in the Default Domain Controllers Policy, but having run the Group Policy Modelling Wizard these settings do not appear to be set for the Domain Controllers OU, should they not be inherited from the Default Domain policy? Additionally, if I link the Default Domain policy to the Domain Controllers OU, the Group Policy Modelling Wizard indicates the expected values for complexity etc, but I still cannot create a new user with my desired password. The domain is running at the Windows Server 2008 R2 functional level. Any thoughts? Thanks! Update: Here is the "Account policy/Password policy" Section from the GPM Wizard: Policy Value Winning GPO Enforce password history 0 Passwords Remembered Default Domain Policy Maximum password age 0 days Default Domain Policy Minimum password age 0 days Default Domain Policy Minimum password length 0 characters Default Domain Policy Passwords must meet complexity Disabled Default Domain Policy These results were taken from running the GPM Wizard at the Domain Controllers OU. I have typed them out by hand as the system I am working on is standalone, this is why the table is not exactly the wording from the Wizard. Are there any other policies that could override the above? Thanks!

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• Aggregating Excel cell contents that match a label [migrated]

  - by Josh

  I'm sure this isn't a terribly difficult thing, but it's not the type of question that easily lends itself to internet searches. I've been assigned a project for work involving a complex spreadsheet. I've done the usual =SUM and other basic Excel formulas, and I've got enough coding background that I'm able to at least fudge my way through VBA, but I'm not certain how to proceed with one part of the task. Simple version: On Sheet 1 I have a list of people (one on each row, person's name in column A), on sheet 2 I have a list of groups (one on each row, group name in column A). Each name in Sheet 1 has its own row, and I have a "Data Validation" dropdown menu where you choose the group each person belongs to. That dropdown is sourced from Sheet 2, where each group has a row. So essentially the data validation source for Sheet 1's "Group" column is just "=Sheet2!$a1:a100" or whatever. The problem is this: I want each group row in Sheet 2 to have a formula which results in a list of all the users which have been assigned to that group on Sheet 1. What I mean is something the equivalent of "select * from PeopleTab where GROUP = ThisGroup". The resulting cell would just stick the names together like "Bob Smith, Joe Jones, Sally Sanderson" I've been Googling for hours but I can't think of a way to phrase my search query to get the results I want. Here's an example of desired result (Dash-delimited. Can't find a way to make it look nice, table tags don't seem to work here): (Sheet 1) Bob Smith - Group 1 (selected from dropdown) Joe Jones - Group 2 (selected from dropdown) Sally Sanderson - Group 1 (selected from dropdown) (Sheet 2) Group 1 - Bob Smith, Sally Sanderson (result of formula) Group 2 - Joe Jones (result of formula) What formula (or even what function) do I use on that second column of sheet 2 to make a flat list out of the members of that group?

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• Junit test bluej [closed]

  - by user1721929

  Can someone make a junit test of this? public class PersonName { int NumberNames(String wholename) { // store the name passed in to the method String testname=wholename; // initialize number of names found int numnames=0; // on each iteration remove one name while(testname.length()>numnames) { // take the "white space" from the beginning and end testname = testname.trim(); // determine the position of the first blank // .. end of the first word int posBlank= testname.indexOf(' '); // cut off word /** * it continues to the stop sign because that is where you commanded it to end */ testname=testname.substring(posBlank+1,testname.length()); // System.out.println(numnames); // System.out.println(testname); numnames++; System.out.println(testname); } return numnames; } public static void main(String args[]) { PersonName One= new PersonName(); System.out.println(One.NumberNames("Bobby")); System.out.println(One.NumberNames("Bobby Smith")); System.out.println(One.NumberNames("Bobby L. Smith")); System.out.println(One.NumberNames(" Bobby Paul Smith Jr. ")); } }

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• Can't install "Docky" theme?

  - by Earl Larson

  So to install a theme for Docky on Ubuntu I had to extract the theme from the Archive Manager via root to /usr/local/share/docky/themes. I did this but it didn't show up in the theme choices. Then I tried just pressing the install button and locating it but I couldn't see the tar file. I then extracted the tar file to the desktop and tried selecting that to install but nothing. What am I doing wrong!?

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• Syncing contacts to iOS device with Exchange

  - by flackend

  I set up a Microsoft Exchange account on my iOS device to sync my Gmail contacts. But Microsoft Exchange is ignoring phone numbers that are labeled as 'iPhone' or 'main'. For example, John Smith: On Mac and Gmail: John Smith main: 123-334-1212 home: 123-330-1002 work: 123-330-8211 iPhone: 123-778-5556 On iOS device (via Exchange sync): John Smith home: 123-330-1002 work: 123-330-8211 I'd like to sync my contacts from my Mac to iCloud and Gmail, but you can't do both: Is there a solution to sync iOS and Gmail contacts without using Exchange? Thanks for any help!

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