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  • Microsoft office communicator 2007 r2 can not send video call to multiple user

    - by mihan007
    so i install ms communication server 2007 r2 and ms office communicator 2007 r2 for three user. so it is possible now to send video call from one person to another. you can use it so way: you choose a person from your contact list and from context menu choose 'send video call'. but when i choose several persons which I want to talk by video this option doesn't exist. (but it should be - http://www.useto.ru/images/01/image015.jpg). what is the problem and how can i tune up communication server or communicator for ability to send video call to several persons?

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  • counting unique values based on multiple columns

    - by gooogalizer
    I am working in google spreadsheets and I am trying to do some counting that takes into consideration cell values across multiple cells in each row. Here's my table: |AUTHOR| |ARTICLE| |VERSION| |PRE-SELECTED| ANDREW GOLF STREAM 1 X ANDREW GOLF STREAM 2 X ANDREW HURRICANES 1 JOHN CAPE COD 1 X JOHN GOLF STREAM 1 (Google doc here) Each person can submit multiple articles as well as multiple versions of the same article. Sometimes different people submit different articles that happen to be identically named (Andrew and John both submitted different articles called "Golf Stream"). Multiple versions written by the same person do not count as unique, but articles with the same title written by different people do count as unique. So, I am looking to find a formula that Counts the number of unique articles that have been submitted [4] (without having to manually create extra columns for doing CONCATS, if possible) It would also be great to find formulas that: Count the number of unique articles that have been pre-selected (marked "X" in "PRE-SELECTED" column) [2] Count the number of unique articles that have only 1 version [4] Count the number of unique articles that have more than 1 of their versions pre-selected 1 Thank you so much! Nikita

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  • Application with two UITabBarControls

    - by kovpas
    Hello, I want to create an app, whith the following structure: UITabBarController ..UINavigationController ....UITabBarController So, in other words, I want to load first UITabbarController as my main ViewController, that appears on application start, in its' first tab there's a UINavigationController with a table, which shows another UITabBarController by pressing on a table cell. I started Tab Based application, set first tab to UINavigationController, and its ViewController to my TableViewController. I also created second UITabBarController in Interface Builder. Created a separate class for this controller. Everything works fine at start - my first UITabBarController is shown perfectly, but when I'm trying to push my second UITabBarController, using my UINavigationController, it looks like nothing is loaded from a .xib file - there are just an empty TabBar and NavigationBar pushed. Seems, that something is wrong with an architecture, but I can't figure out the problem. Thanks in advance and I'm sorry for my poor English. Please let me know if something is not clear, I'll try to rephrase it :).

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  • Make list from two database items

    - by unnamet
    I want to make a list from some data that I have in my database. The first two sets of data in my database are first name and last name. I want my list to show both first and last name instead of now where it only shows the first name. How do I do that? My code looks like this: private void fillData() { Cursor contactCursor = mDbHelper.fetchAllReminders(); startManagingCursor(contactCursor); String[] from = new String[]{DbAdapter.KEY_FIRST}; int[] to = new int[]{R.id.contactlist}; SimpleCursorAdapter contacts = new SimpleCursorAdapter(this, R.layout.list, contactCursor, from, to); setListAdapter(contacts); }

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  • How to count each digit in a range of integers?

    - by Carlos Gutiérrez
    Imagine you sell those metallic digits used to number houses, locker doors, hotel rooms, etc. You need to find how many of each digit to ship when your customer needs to number doors/houses: 1 to 100 51 to 300 1 to 2,000 with zeros to the left The obvious solution is to do a loop from the first to the last number, convert the counter to a string with or without zeros to the left, extract each digit and use it as an index to increment an array of 10 integers. I wonder if there is a better way to solve this, without having to loop through the entire integers range. Solutions in any language or pseudocode are welcome. Edit: Answers review John at CashCommons and Wayne Conrad comment that my current approach is good and fast enough. Let me use a silly analogy: If you were given the task of counting the squares in a chess board in less than 1 minute, you could finish the task by counting the squares one by one, but a better solution is to count the sides and do a multiplication, because you later may be asked to count the tiles in a building. Alex Reisner points to a very interesting mathematical law that, unfortunately, doesn’t seem to be relevant to this problem. Andres suggests the same algorithm I’m using, but extracting digits with %10 operations instead of substrings. John at CashCommons and phord propose pre-calculating the digits required and storing them in a lookup table or, for raw speed, an array. This could be a good solution if we had an absolute, unmovable, set in stone, maximum integer value. I’ve never seen one of those. High-Performance Mark and strainer computed the needed digits for various ranges. The result for one millon seems to indicate there is a proportion, but the results for other number show different proportions. strainer found some formulas that may be used to count digit for number which are a power of ten. Robert Harvey had a very interesting experience posting the question at MathOverflow. One of the math guys wrote a solution using mathematical notation. Aaronaught developed and tested a solution using mathematics. After posting it he reviewed the formulas originated from Math Overflow and found a flaw in it (point to Stackoverflow :). noahlavine developed an algorithm and presented it in pseudocode. A new solution After reading all the answers, and doing some experiments, I found that for a range of integer from 1 to 10n-1: For digits 1 to 9, n*10(n-1) pieces are needed For digit 0, if not using leading zeros, n*10n-1 - ((10n-1) / 9) are needed For digit 0, if using leading zeros, n*10n-1 - n are needed The first formula was found by strainer (and probably by others), and I found the other two by trial and error (but they may be included in other answers). For example, if n = 6, range is 1 to 999,999: For digits 1 to 9 we need 6*105 = 600,000 of each one For digit 0, without leading zeros, we need 6*105 – (106-1)/9 = 600,000 - 111,111 = 488,889 For digit 0, with leading zeros, we need 6*105 – 6 = 599,994 These numbers can be checked using High-Performance Mark results. Using these formulas, I improved the original algorithm. It still loops from the first to the last number in the range of integers, but, if it finds a number which is a power of ten, it uses the formulas to add to the digits count the quantity for a full range of 1 to 9 or 1 to 99 or 1 to 999 etc. Here's the algorithm in pseudocode: integer First,Last //First and last number in the range integer Number //Current number in the loop integer Power //Power is the n in 10^n in the formulas integer Nines //Nines is the resut of 10^n - 1, 10^5 - 1 = 99999 integer Prefix //First digits in a number. For 14,200, prefix is 142 array 0..9 Digits //Will hold the count for all the digits FOR Number = First TO Last CALL TallyDigitsForOneNumber WITH Number,1 //Tally the count of each digit //in the number, increment by 1 //Start of optimization. Comments are for Number = 1,000 and Last = 8,000. Power = Zeros at the end of number //For 1,000, Power = 3 IF Power 0 //The number ends in 0 00 000 etc Nines = 10^Power-1 //Nines = 10^3 - 1 = 1000 - 1 = 999 IF Number+Nines <= Last //If 1,000+999 < 8,000, add a full set Digits[0-9] += Power*10^(Power-1) //Add 3*10^(3-1) = 300 to digits 0 to 9 Digits[0] -= -Power //Adjust digit 0 (leading zeros formula) Prefix = First digits of Number //For 1000, prefix is 1 CALL TallyDigitsForOneNumber WITH Prefix,Nines //Tally the count of each //digit in prefix, //increment by 999 Number += Nines //Increment the loop counter 999 cycles ENDIF ENDIF //End of optimization ENDFOR SUBROUTINE TallyDigitsForOneNumber PARAMS Number,Count REPEAT Digits [ Number % 10 ] += Count Number = Number / 10 UNTIL Number = 0 For example, for range 786 to 3,021, the counter will be incremented: By 1 from 786 to 790 (5 cycles) By 9 from 790 to 799 (1 cycle) By 1 from 799 to 800 By 99 from 800 to 899 By 1 from 899 to 900 By 99 from 900 to 999 By 1 from 999 to 1000 By 999 from 1000 to 1999 By 1 from 1999 to 2000 By 999 from 2000 to 2999 By 1 from 2999 to 3000 By 1 from 3000 to 3010 (10 cycles) By 9 from 3010 to 3019 (1 cycle) By 1 from 3019 to 3021 (2 cycles) Total: 28 cycles Without optimization: 2,235 cycles Note that this algorithm solves the problem without leading zeros. To use it with leading zeros, I used a hack: If range 700 to 1,000 with leading zeros is needed, use the algorithm for 10,700 to 11,000 and then substract 1,000 - 700 = 300 from the count of digit 1. Benchmark and Source code I tested the original approach, the same approach using %10 and the new solution for some large ranges, with these results: Original 104.78 seconds With %10 83.66 With Powers of Ten 0.07 A screenshot of the benchmark application: If you would like to see the full source code or run the benchmark, use these links: Complete Source code (in Clarion): http://sca.mx/ftp/countdigits.txt Compilable project and win32 exe: http://sca.mx/ftp/countdigits.zip Accepted answer noahlavine solution may be correct, but l just couldn’t follow the pseudo code, I think there are some details missing or not completely explained. Aaronaught solution seems to be correct, but the code is just too complex for my taste. I accepted strainer’s answer, because his line of thought guided me to develop this new solution.

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  • VBA-Sorting the data in a listbox, sort works but data in listbox not changed

    - by Mike Clemens
    A listbox is passed, the data placed in an array, the array is sort and then the data is placed back in the listbox. The part that does work is putting the data back in the listbox. Its like the listbox is being passed by value instead of by ref. Here's the sub that does the sort and the line of code that calls the sort sub. Private Sub SortListBox(ByRef LB As MSForms.ListBox) Dim First As Integer Dim Last As Integer Dim NumItems As Integer Dim i As Integer Dim j As Integer Dim Temp As String Dim TempArray() As Variant ReDim TempArray(LB.ListCount) First = LBound(TempArray) ' this works correctly Last = UBound(TempArray) - 1 ' this works correctly For i = First To Last TempArray(i) = LB.List(i) ' this works correctly Next i For i = First To Last For j = i + 1 To Last If TempArray(i) > TempArray(j) Then Temp = TempArray(j) TempArray(j) = TempArray(i) TempArray(i) = Temp End If Next j Next i ! data is now sorted LB.Clear ! this doesn't clear the items in the listbox For i = First To Last LB.AddItem TempArray(i) ! this doesn't work either Next i End Sub Private Sub InitializeForm() ' There's code here to put data in the list box Call SortListBox(FieldSelect.CompleteList) End Sub Thanks for your help.

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  • JSF: How to get the selected item from selectOneMenu if its rendering is dynamic?

    - by Dzmitry Zhaleznichenka
    At my view I have two menus that I want to make dependent, namely, if first menu holds values "render second menu" and "don't render second menu", I want second menu to be rendered only if user selects "render second menu" option in the first menu. After second menu renders at the same page as the first one, user has to select current item from it, fill another fields and submit the form to store values in database. Both the lists of options are static, they are obtained from the database once and stay the same. My problem is I always get null as a value of the selected item from second menu. How to get a proper value? The sample code of view that holds problematic elements is: <h:selectOneMenu id="renderSecond" value="#{Bean.renderSecondValue}" valueChangeListener="#{Bean.updateDependentMenus}" immediate="true" onchange="this.form.submit();" > <f:selectItems value="#{Bean.typesOfRendering}" /> </h:selectOneMenu><br /> <h:selectOneMenu id="iWillReturnYouZeroAnyway" value="#{Bean.currentItem}" rendered="#{Bean.rendered}" > <f:selectItems value="#{Bean.items}" /> </h:selectOneMenu><br /> <h:commandButton action="#{Bean.store}" value="#Store" /> However, if I remove "rendered" attribute from the second menu, everything works properly, except for displaying the menu for all the time that I try to prevent, so I suggest the problem is in behavior of dynamic rendering. The initial value of isRendered is false because the default item in first menu is "don't render second menu". When I change value in first menu and update isRendered with valueChangeListener, the second menu displays but it doesn't initialize currentItem while submitting. Some code from my backing bean is below: public void updateDependentMenus(ValueChangeEvent value) { String newValue = (String) value.getNewValue(); if ("render second menu" == newValue){ isRendered = true; } else { isRendered = false; } } public String store(){ System.out.println(currentItem); return "stored"; }

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  • IIS - Forwarding requests to a folder to another port

    - by user1231958
    Context I currently installed Glassfish 3 in a server that currently holds ASP and PHP inside Internet Information Server 7 so we can start moving to a new system architecture (the information system is being remade). Obviously, Glassfish uses another port and without too much configuration (all I had to do is to install it) it worked. If I write www.domain.com:8080, the person will be redirected to the Glassfish server. Issue Obviously I don't want the person to write the port! I also believe it might also hold some security issues. Requirement I need the server to take an address of the form www.domain.com/gf or new.domain.com or something alike, and when it receives such a request, "redirect" (masking the URL) the user to the Glassfish website (www.domain.com:8080). Thank you beforehand!

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  • FBConnect for iPhone: sessionDidNotLogin, sessionDidLogout, session didLogin not called the second t

    - by Irene
    My problem is very similar to this question, however I am posting a new one, as the answer to the aforementioned does not seem to solve my problem. I have a multiview application - the first view is where the user logs in to Facebook, and the second where he picks an image and uploads it there. The first time the app runs, everything works fine, however if I return to the login view and press logout, then any calls to sessionDidNotLogin, sessionDidLogout or session didLogin don't seem to work. I found out that the first time, if I NSLog(@"%@",session.delegates); I have 2; my LoginViewController and the FBLoginButton. However, apart from that first time, the same log prints only the LoginViewController and not the FBLoginButton. I guess this is connected somehow, but I don't know how to solve it. Do I have to manually add the FBLoginButton to the session delegates, or I'm doing something else wrong here? Thank you for any help/suggestion.

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  • Sending files with iChat doesnt work in Snow Leopard

    - by Nippysaurus
    My girlfriend and I often sit in the lounge looking at stuff online and sending it to each other to look at. So sometimes we drag images into an iChat and it will appear in the other persons iChat window. The ability to do this seems to have broken with the upgrade to 10.6, and the issue persists over both of our laptops and my iMac. The following is a scheenshot of what happens. The image that appears in my window is from me dragging it in there, but the other person cannot see it, and the next bubble is the other person attempting to send me an image, but I only ever see that little white circle for a while then it dissapears. Is this broken for everyone?

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  • Polynomial division overloading operator (solved)

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) const { polinom Result, temp2; polinom temp = *this; Iter i = temp.poly.begin(); constIter j = P.poly.begin(); int resultSize = 0; if (temp.poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); temp = temp - Result * P; } else { Result.insert(0, 0); } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); if (Result.poly.size() < 2) temp2 = Result; else { temp2 = Result; resultSize = Result.poly.size(); for (int k = 1 ; k != resultSize; k++) temp2.poly.pop_front(); } temp = temp - temp2 * P; } else break; } } return Result; } }; The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Final Update After listening to Dave, I finally made it by overloading both / and & to return the quotient and the remainder so thanks a lot everyone for your help and especially you Dave for your great idea! P.S. If anyone wants for me to post these 2 overloaded operator please ask it by commenting on my post (and maybe give a vote up for everyone involved).

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  • Skype Optimization - Port Forwarding on a Router

    - by user19185
    I was watching this Video which talked about using port-forwarding to optimize your LAN for skype calls. According to the video, as explained in the first couple of minutes in the video, the reason you would need optimization is because if the person your call has a firewall setup, your connection has to go-through a third-party computer to connect to them. I believe I stated this correct (maybe not). None the less, my question is this: do both parties on the call need to enable port forwarding to optimize skype, or just one party (person)?

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  • Programmatically controlling a Dojo Accordion

    - by prule
    I have a dijit.layout.AccordionContainer on my page which is defined in the html and created when dojo parses the page on load. Then, as the user interacts with the page I use Ajax to retrieve data and programmatically populate the container (removing existing items first). To illustrate my issue simply, here is some code that doesn't work: function doit() { var accordion = dijit.byId("accordionShell"); accordion.getChildren().each(function(item) { accordion.removeChild(item); }); for (i = 1; i < 5; i++) { var d = new dijit.layout.AccordionPane({title:'hello', content:'world'}); accordion.addChild(d); } } This fails, because only the first item in the accordian is visible. I think the others actually exist, but they are not visible so you can't do anything. I've managed to get around it by: Always ensuring there is 1 item in the accordian (so I never remove the first child) Call accordian.layout() after changing the contents So, this code "works" as long as you always want to see the first item, and don't actually expand any but the first one: function doit() { var accordion = dijit.byId("accordionShell"); var i = 0; accordion.getChildren().each(function(item) { if (i > 0) accordion.removeChild(item); i++; }); for (i = 1; i < 5; i++) { var d = new dijit.layout.AccordionPane({title:'hello', content:'world'}); accordion.addChild(d); } accordion.layout(); } I am using Dojo 1.2.0 - Anyone know what I am doing wrong?

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  • Disable Memory Modules In BIOS for Testing Purposes (Optimize Nehalem/Gulftown Memory Performance)

    - by Bob
    I recently acquired an HP Z800 with two Intel Xeon X5650 (Gulftown) 6 core processors. The person that configured the system chose 16GB (8 x 2GB DDR3-1333). I'm assuming this person was unaware these processors have 3 memory channels and to optimize memory performance one should choose memory in multiples of three. Based on this information, I have a question: By entering the BIOS, can I disable the bank on each processor that has the single memory module? If so, will this have any adverse effects or behave differently than physically removing the modules? I ask due to the fact that I prefer to store the extra memory in the system if it truly behaves as if the memory is not even there. Also, I see this as an opportunity to test 12GB vs. 16GB to see if there is a noticeable difference. Note: According to http://www.delltechcenter.com/page/04-08-2009+-+Nehalem+and+Memory+Configurations?t=anon, the current configuration reduces the overall data transfer speed to 1066 and in addition, the memory bandwidth goes down by about 23%.

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  • Jquery runtime error: object expected

    - by Joris
    The Jquery script that controls my tabcontainer gives an "object expected" runtime error. I honestly can't find the reason why: $(document).ready(function() { //When page loads... $(".tab_content").hide(); //Hide all content $("ul.tabs li:first").addClass("active").show(); //Activate first tab $(".tab_content:first").show(); //Show first tab content //On Click Event $("ul.tabs li").click(function() { $("ul.tabs li").removeClass("active"); //Remove any "active" class $(this).addClass("active"); //Add "active" class to selected tab $(".tab_content").hide(); //Hide all tab content var activeTab = $(this).find("a").attr("href"); //Find the href attribute value to identify the active tab + content $(activeTab).fadeIn(); //Fade in the active ID content return false; }); }); Has it something to do with the stylesheet?

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  • iphone SDK Keyboard problem

    - by Elishiah
    I have a login view controller that has a button and a UI text view that becomes the first responder. When the button is clicked it loads the forgot view controller and a new UI textview becomes the first responder. When I dissmis the modal view controller the keyboard also gets dismissed and when I load the forgot view controller again the keyboard will not pop up as the first responder. (IBAction) backBeg: (id) sender{ [self dismissModalViewControllerAnimated:YES]; }

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  • Office communicator voice chat

    - by Gareth Simpson
    My company recently mandated a switch from Skype to Office Communicator for IM / voice chat. While Skype was never the be all and end all of VOIP, it was at least usable. With Communicator, if one person is talking, everyone else is basically silent (or as good as) so a normal conversation is impossible. No one can interrupt anyone else, if two people start talking at once it's a crap shoot who gets to be heard and often the person who is inaudible doesn't know it. There do not seem to be any client side settings to fix this. Is there anything that can be done server side or is it just rubbish?

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  • Two virtual computers using one laptop?

    - by JP19
    Hi, I have a laptop and an external display. I am able to use the external display either as extended desktop OR duplicate desktop. I also use a separate keyboard and mouse. So effectively, the laptop acts just as a box without display. I want to know is it possible to use the laptop as 2 virtual PCs. (with separate control). I.E., one person uses the laptop as a laptop, and other person can simultaneously connect a keyboard and mouse and the external display and effectively get a second PC. OS I am using - windows (thought solution for other OSes are welcome). Laptop is a Dell Precision M6300. thanks JP

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  • Oracle data warehouse design - fact table acting as a dimension?

    - by Elizabeth
    THANKS: Both answers here are very helpful, but I could only pick one. I really appreciate the advice! our datawarehouse will be used more for workflow reports than traditional analytical reports. Our users care about "current picture" far more than history. (though history matters, too.) We are a government entity that does not have costs or related calculations. Mostly just counts of people within given locations and with related history. We are using Oracle, and I have found distinct advantage in using the star join whenever possible and would like to rearchitect everything to as closely resemble the star schema as is reasonable for our business uses. Speed in this DW is vital, and a number of tests have already proven the star schema approach to me. Our "person" table is key - it contains over 4 million records and will be the most frequently used source in queries. It can be seen at the center of a star with multiple dimensions (like age, gender, affiliation, location, etc.). It is a very LONG table, particularly when I join it to the address and contact information. However, it is more like a dimension table when we start looking at history. For example, there are two different history tables that have a person key pointing to the person table. One has over 20 million records and the other has almost 50 million and grows daily. Is this table a fact table or a dimension table? Can one work as both? If so, is that going to be a big performance problem? Is it common to query more off of a dimension than a fact? What happens if a DIFFERENT fact table that uses the person table as a dimension is actually only 60,000 records (much smaller.). I think my problem is that our data and use of it does not fit with the commonly use examples of star schemas. CLARIFICATION: Some good thoughts have been added below, but perhaps I left too much out to really explain well. Here's some more info: We handle a voter database. We don't have any measures except voter counts by various groups: voter counts by party, by age, by location; voter counts by ballot type and election, by ballot status and election, etc. We do have a "voting history" log as well as an activity audit log (change of address, party, etc.). We have information on which voters are election workers and all that related information. I figure I'll get to the peripheral stuff later. For now I'm focusing on our two major "business processes": voter registration(which IS a voter.) and election turnout. In the first, voter is a fact. In the second, voter is a dimension, along with party, election, and type of ballot. (and in case anyone is worried - no we don't know HOW people vote. Just that they do. LOL ) I hope that clarifies things a bit.

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  • How to get form elements from one form to another form in drupal

    - by ugesh.gali
    I have two forms in drupal and the form values is stores in DB.The first form contanins some fields related to company by default like id conpany,name etc.The second form consists of attributes .If I add attributes to the second form the attribute field should be visible in first form. Ex If i add loaction or any other field it should visible in first form as location field.

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  • jqgrid ASP.NET MVC -- getting data right for the grid.

    - by SamM09
    Here is my dilemma, I have not been able to manipulate my data to a form fitting to jqgrid standards. This is my first time using the jqgrid and I've spent a lot of time reading up on it. My js code is as follows: jQuery("#list").jqGrid({ url: '/Home/ListContacts/', dataType: "json", contentType: "application/json; charset=utf-8", mtype: 'POST', colNames: ['First Name', 'MI', 'Last Name'], colModel: [ { name: 'First Name', index: 'FName', width: 40, align: 'left' }, { name: 'MI', index: 'MInitial', width: 40, align: 'left' }, { name: 'Last Name', index: 'LName', width: 400, align: 'left'}], pager: jQuery('#pager'), rowNum: 10, rowList: [5, 10, 20, 50], sortname: 'Id', sortorder: "desc", repeatitems: false, viewrecords: true, imgpath: '/scripts/themes/basic/images', caption: 'My first grid' }); }); what im getting from the database: [["4","Jenna","Mccarthy"],["56","wer","weoiru"]] Now correct me if I am wrong, but the index: in my colModel refers to the column names in my database right? Could someone point to a reference that is straight forward or just start me off with this I would be most grateful.

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  • codingBat separateThousands using regex (and unit testing how-to)

    - by polygenelubricants
    This question is a combination of regex practice and unit testing practice. Regex part I authored this problem separateThousands for personal practice: Given a number as a string, introduce commas to separate thousands. The number may contain an optional minus sign, and an optional decimal part. There will not be any superfluous leading zeroes. Here's my solution: String separateThousands(String s) { return s.replaceAll( String.format("(?:%s)|(?:%s)", "(?<=\\G\\d{3})(?=\\d)", "(?<=^-?\\d{1,3})(?=(?:\\d{3})+(?!\\d))" ), "," ); } The way it works is that it classifies two types of commas, the first, and the rest. In the above regex, the rest subpattern actually appears before the first. A match will always be zero-length, which will be replaceAll with ",". The rest basically looks behind to see if there was a match followed by 3 digits, and looks ahead to see if there's a digit. It's some sort of a chain reaction mechanism triggered by the previous match. The first basically looks behind for ^ anchor, followed by an optional minus sign, and between 1 to 3 digits. The rest of the string from that point must match triplets of digits, followed by a nondigit (which could either be $ or \.). My question for this part is: Can this regex be simplified? Can it be optimized further? Ordering rest before first is deliberate, since first is only needed once No capturing group Unit testing part As I've mentioned, I'm the author of this problem, so I'm also the one responsible for coming up with testcases for them. Here they are: INPUT, OUTPUT "1000", "1,000" "-12345", "-12,345" "-1234567890.1234567890", "-1,234,567,890.1234567890" "123.456", "123.456" ".666666", ".666666" "0", "0" "123456789", "123,456,789" "1234.5678", "1,234.5678" "-55555.55555", "-55,555.55555" "0.123456789", "0.123456789" "123456.789", "123,456.789" I haven't had much experience with industrial-strength unit testing, so I'm wondering if others can comment whether this is a good coverage, whether I've missed anything important, etc (I can always add more tests if there's a scenario I've missed).

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  • Binary Cosine Cofficient

    - by hairyyak
    I was given the following forumulae for calculating this sim=|QnD| / v|Q|v|D| I went ahed and implemented a class to compare strings consisting of a series of words #pragma once #include <vector> #include <string> #include <iostream> #include <vector> using namespace std; class StringSet { public: StringSet(void); StringSet( const string the_strings[], const int no_of_strings); ~StringSet(void); StringSet( const vector<string> the_strings); void add_string( const string the_string); bool remove_string( const string the_string); void clear_set(void); int no_of_strings(void) const; friend ostream& operator <<(ostream& outs, StringSet& the_strings); friend StringSet operator *(const StringSet& first, const StringSet& second); friend StringSet operator +(const StringSet& first, const StringSet& second); double binary_coefficient( const StringSet& the_second_set); private: vector<string> set; }; #include "StdAfx.h" #include "StringSet.h" #include <iterator> #include <algorithm> #include <stdexcept> #include <iostream> #include <cmath> StringSet::StringSet(void) { } StringSet::~StringSet(void) { } StringSet::StringSet( const vector<string> the_strings) { set = the_strings; } StringSet::StringSet( const string the_strings[], const int no_of_strings) { copy( the_strings, &the_strings[no_of_strings], back_inserter(set)); } void StringSet::add_string( const string the_string) { try { if( find( set.begin(), set.end(), the_string) == set.end()) { set.push_back(the_string); } else { //String is already in the set. throw domain_error("String is already in the set"); } } catch( domain_error e) { cout << e.what(); exit(1); } } bool StringSet::remove_string( const string the_string) { //Found the occurrence of the string. return it an iterator pointing to it. vector<string>::iterator iter; if( ( iter = find( set.begin(), set.end(), the_string) ) != set.end()) { set.erase(iter); return true; } return false; } void StringSet::clear_set(void) { set.clear(); } int StringSet::no_of_strings(void) const { return set.size(); } ostream& operator <<(ostream& outs, StringSet& the_strings) { vector<string>::const_iterator const_iter = the_strings.set.begin(); for( ; const_iter != the_strings.set.end(); const_iter++) { cout << *const_iter << " "; } cout << endl; return outs; } //This function returns the union of the two string sets. StringSet operator *(const StringSet& first, const StringSet& second) { vector<string> new_string_set; new_string_set = first.set; for( unsigned int i = 0; i < second.set.size(); i++) { vector<string>::const_iterator const_iter = find(new_string_set.begin(), new_string_set.end(), second.set[i]); //String is new - include it. if( const_iter == new_string_set.end() ) { new_string_set.push_back(second.set[i]); } } StringSet the_set(new_string_set); return the_set; } //This method returns the intersection of the two string sets. StringSet operator +(const StringSet& first, const StringSet& second) { //For each string in the first string look though the second and see if //there is a matching pair, in which case include the string in the set. vector<string> new_string_set; vector<string>::const_iterator const_iter = first.set.begin(); for ( ; const_iter != first.set.end(); ++const_iter) { //Then search through the entire second string to see if //there is a duplicate. vector<string>::const_iterator const_iter2 = second.set.begin(); for( ; const_iter2 != second.set.end(); const_iter2++) { if( *const_iter == *const_iter2 ) { new_string_set.push_back(*const_iter); } } } StringSet new_set(new_string_set); return new_set; } double StringSet::binary_coefficient( const StringSet& the_second_set) { double coefficient; StringSet intersection = the_second_set + set; coefficient = intersection.no_of_strings() / sqrt((double) no_of_strings()) * sqrt((double)the_second_set.no_of_strings()); return coefficient; } However when I try and calculate the coefficient using the following main function: // Exercise13.cpp : main project file. #include "stdafx.h" #include <boost/regex.hpp> #include "StringSet.h" using namespace System; using namespace System::Runtime::InteropServices; using namespace boost; //This function takes as input a string, which //is then broken down into a series of words //where the punctuaction is ignored. StringSet break_string( const string the_string) { regex re; cmatch matches; StringSet words; string search_pattern = "\\b(\\w)+\\b"; try { // Assign the regular expression for parsing. re = search_pattern; } catch( regex_error& e) { cout << search_pattern << " is not a valid regular expression: \"" << e.what() << "\"" << endl; exit(1); } sregex_token_iterator p(the_string.begin(), the_string.end(), re, 0); sregex_token_iterator end; for( ; p != end; ++p) { string new_string(p->first, p->second); String^ copy_han = gcnew String(new_string.c_str()); String^ copy_han2 = copy_han->ToLower(); char* str2 = (char*)(void*)Marshal::StringToHGlobalAnsi(copy_han2); string new_string2(str2); words.add_string(new_string2); } return words; } int main(array<System::String ^> ^args) { StringSet words = break_string("Here is a string, with some; words"); StringSet words2 = break_string("There is another string,"); cout << words.binary_coefficient(words2); return 0; } I get an index which is 1.5116 rather than a value from 0 to 1. Does anybody have a clue why this is the case? Any help would be appreciated.

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  • Django QuerySet API: How do I join iexact and icontains?

    - by Zeynel
    Hello, I have this join: lawyers = Lawyer.objects.filter(last__iexact=last_name).filter(first__icontains=first_name) This is the site If you try Last Name: Abbas and First Name: Amr it tells you that amr abbas has 1 schoolmates. But if you try First name only it says that there are no lawyers in the database called amr (obviously there is). If I change (last__iexact=last_name) to (last__icontains=last_name) then leaving Last Name blank works fine and amr is found. But with last__icontains=last_name if you search for "collin" you also get "collins" and "collingwood" which is not what I want. Do you know how I can use iexact and also have it ignored if it is blank? Thanks This is the view function: def search_form(request): if request.method == 'POST': search_form = SearchForm(request.POST) if search_form.is_valid(): last_name = search_form.cleaned_data['last_name'] first_name = search_form.cleaned_data['first_name'] lawyers = Lawyer.objects.filter(last__iexact=last_name).filter(first__icontains=first_name) if len(lawyers)==0: form = SearchForm() return render_to_response('not_in_database.html', {'last': last_name, 'first': first_name, 'form': form}) if len(lawyers)>1: form = SearchForm(initial={'last_name': last_name}) return render_to_response('more_than_1_match.html', {'lawyers': lawyers, 'last': last_name, 'first': first_name, 'form': form}) q_school = Lawyer.objects.filter(last__icontains=last_name).filter(first__icontains=first_name).values_list('school', flat=True) q_year = Lawyer.objects.filter(last__icontains=last_name).filter(first__icontains=first_name).values_list('year_graduated', flat=True) lawyers1 = Lawyer.objects.filter(school__iexact=q_school[0]).filter(year_graduated__icontains=q_year[0]).exclude(last__icontains=last_name) form = SearchForm() return render_to_response('search_results.html', {'lawyers': lawyers1, 'last': last_name, 'first': first_name, 'form': form}) else: form = SearchForm() return render_to_response('search_form.html', {'form': form, })

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  • Zend RegEx Validator error message issue

    - by Mallika Iyer
    Hello, I'm validating a text field in my form as follows: $name = new Zend_Form_Element_Text('name'); $name->setLabel('First Name:') ->setRequired(true) ->addFilter(new Zend_Filter_StringTrim()) ->addValidator('regex',true,array('/^[(a-zA-Z0-9)]+$/')) ->addErrorMessage('Please enter a valid first name'); What I'm trying to accomplish is - how can i display a meaningful error message? Eg: If first name is 'XYZ-', how can i display '- is not allowed in first name.' Is there a way I can access what character the regex is failing for? Would you recommend something else altogether? I thought about writing a custom validator but the regex is pretty simple, so I don't see the point. I couldn't find a decent documentation for the zend 'regex' validator anywhere. If I don't override the default error message, I simple get something like : ';;;hhbhbhb' does not match against pattern '/^[(a-zA-Z0-9)]+$/' - which I obviously don't want to display to the user. I'd appreciate your inputs.

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