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  • PHP: Display comma after each element except the last. Using 'for' statement and no 'implode/explode

    - by Jonathan
    hi, so I have this simple for loop to echo an array: for ($i = 0; $i < count($director); $i++) { echo '<a href="person.php?id='.$director[$i]["id"].'">'.$director[$i]["name"].'</a>'; } The problem here is that when more than one element is in the array then I get everything echoed without any space between. I want to separate each element with a comma except the last one. I cant use 'implode' so I'm looking for another solution... Anyone?

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  • GWT Editor: How to set last modified time on the entity when saved?

    - by Mike
    Suppose at client side i have an Entity proxy to edit by the UI and when i click save button, the last modified time is save in the entity as a field. //start MyEntityProxy proxy = getProxy();//fetched from server Request<Void> saveRequest = requestFact.myEntityProxyRequest().save(proxy); editorDriver.edit(proxy, saveRequest.getRequestContext()); editorDriver.flush(); //user modifies UI .... //save editorDriver.flush(); saveRequest.fire(); The problem is, where to insert the proxy.setLastModifiedTime(data) call? I always got java.lang.IllegalStateException: The AutoBean has been frozen. Thanks.

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  • "Remember" last three MySql queries; Cookie, passed variable or other method?

    - by Camran
    I have a classified website, with pretty sophisticated searching, and I am about to implement a function where the last three queries is displayed for the user, so that the user can go back easier through the queries. This because for each query the user has to provide a lot of input. I have four questions for you: I wonder, how can I save the actual query (SELECT * FROM etc etc)...? Do I need to add some form of encryption to be on the safe side? How will this affect performance? (I don't like the fact that cookies slow websites down) Anything else to think about? If you need more input, let me know... Btw, the website is PHP based. Thanks

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  • Get the folder where the last mailitem was moved in Outlook?

    - by user2892971
    I have a vbscript macro that I'm using in Outlook. It moves a mailitem to some folder, say X. After I run the macro and I try to manually move a mailitem from Outlook with Control-v, it defaults to folder X. I would like Control-v to default to the folder that it would have used before I ran the macro. Is there some way in VBScript to find out what folder the last mailitem was move to, and to return that to be the default folder after I run my script? Or is there a way to move a mailitem in my script without the destination folder being remembered by Outlook Control-v after I run the script? Thanks for any hints.

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  • Parallelism in .NET – Part 11, Divide and Conquer via Parallel.Invoke

    - by Reed
    Many algorithms are easily written to work via recursion.  For example, most data-oriented tasks where a tree of data must be processed are much more easily handled by starting at the root, and recursively “walking” the tree.  Some algorithms work this way on flat data structures, such as arrays, as well.  This is a form of divide and conquer: an algorithm design which is based around breaking up a set of work recursively, “dividing” the total work in each recursive step, and “conquering” the work when the remaining work is small enough to be solved easily. Recursive algorithms, especially ones based on a form of divide and conquer, are often a very good candidate for parallelization. This is apparent from a common sense standpoint.  Since we’re dividing up the total work in the algorithm, we have an obvious, built-in partitioning scheme.  Once partitioned, the data can be worked upon independently, so there is good, clean isolation of data. Implementing this type of algorithm is fairly simple.  The Parallel class in .NET 4 includes a method suited for this type of operation: Parallel.Invoke.  This method works by taking any number of delegates defined as an Action, and operating them all in parallel.  The method returns when every delegate has completed: Parallel.Invoke( () => { Console.WriteLine("Action 1 executing in thread {0}", Thread.CurrentThread.ManagedThreadId); }, () => { Console.WriteLine("Action 2 executing in thread {0}", Thread.CurrentThread.ManagedThreadId); }, () => { Console.WriteLine("Action 3 executing in thread {0}", Thread.CurrentThread.ManagedThreadId); } ); .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; } Running this simple example demonstrates the ease of using this method.  For example, on my system, I get three separate thread IDs when running the above code.  By allowing any number of delegates to be executed directly, concurrently, the Parallel.Invoke method provides us an easy way to parallelize any algorithm based on divide and conquer.  We can divide our work in each step, and execute each task in parallel, recursively. For example, suppose we wanted to implement our own quicksort routine.  The quicksort algorithm can be designed based on divide and conquer.  In each iteration, we pick a pivot point, and use that to partition the total array.  We swap the elements around the pivot, then recursively sort the lists on each side of the pivot.  For example, let’s look at this simple, sequential implementation of quicksort: public static void QuickSort<T>(T[] array) where T : IComparable<T> { QuickSortInternal(array, 0, array.Length - 1); } private static void QuickSortInternal<T>(T[] array, int left, int right) where T : IComparable<T> { if (left >= right) { return; } SwapElements(array, left, (left + right) / 2); int last = left; for (int current = left + 1; current <= right; ++current) { if (array[current].CompareTo(array[left]) < 0) { ++last; SwapElements(array, last, current); } } SwapElements(array, left, last); QuickSortInternal(array, left, last - 1); QuickSortInternal(array, last + 1, right); } static void SwapElements<T>(T[] array, int i, int j) { T temp = array[i]; array[i] = array[j]; array[j] = temp; } Here, we implement the quicksort algorithm in a very common, divide and conquer approach.  Running this against the built-in Array.Sort routine shows that we get the exact same answers (although the framework’s sort routine is slightly faster).  On my system, for example, I can use framework’s sort to sort ten million random doubles in about 7.3s, and this implementation takes about 9.3s on average. Looking at this routine, though, there is a clear opportunity to parallelize.  At the end of QuickSortInternal, we recursively call into QuickSortInternal with each partition of the array after the pivot is chosen.  This can be rewritten to use Parallel.Invoke by simply changing it to: // Code above is unchanged... SwapElements(array, left, last); Parallel.Invoke( () => QuickSortInternal(array, left, last - 1), () => QuickSortInternal(array, last + 1, right) ); } This routine will now run in parallel.  When executing, we now see the CPU usage across all cores spike while it executes.  However, there is a significant problem here – by parallelizing this routine, we took it from an execution time of 9.3s to an execution time of approximately 14 seconds!  We’re using more resources as seen in the CPU usage, but the overall result is a dramatic slowdown in overall processing time. This occurs because parallelization adds overhead.  Each time we split this array, we spawn two new tasks to parallelize this algorithm!  This is far, far too many tasks for our cores to operate upon at a single time.  In effect, we’re “over-parallelizing” this routine.  This is a common problem when working with divide and conquer algorithms, and leads to an important observation: When parallelizing a recursive routine, take special care not to add more tasks than necessary to fully utilize your system. This can be done with a few different approaches, in this case.  Typically, the way to handle this is to stop parallelizing the routine at a certain point, and revert back to the serial approach.  Since the first few recursions will all still be parallelized, our “deeper” recursive tasks will be running in parallel, and can take full advantage of the machine.  This also dramatically reduces the overhead added by parallelizing, since we’re only adding overhead for the first few recursive calls.  There are two basic approaches we can take here.  The first approach would be to look at the total work size, and if it’s smaller than a specific threshold, revert to our serial implementation.  In this case, we could just check right-left, and if it’s under a threshold, call the methods directly instead of using Parallel.Invoke. The second approach is to track how “deep” in the “tree” we are currently at, and if we are below some number of levels, stop parallelizing.  This approach is a more general-purpose approach, since it works on routines which parse trees as well as routines working off of a single array, but may not work as well if a poor partitioning strategy is chosen or the tree is not balanced evenly. This can be written very easily.  If we pass a maxDepth parameter into our internal routine, we can restrict the amount of times we parallelize by changing the recursive call to: // Code above is unchanged... SwapElements(array, left, last); if (maxDepth < 1) { QuickSortInternal(array, left, last - 1, maxDepth); QuickSortInternal(array, last + 1, right, maxDepth); } else { --maxDepth; Parallel.Invoke( () => QuickSortInternal(array, left, last - 1, maxDepth), () => QuickSortInternal(array, last + 1, right, maxDepth)); } We no longer allow this to parallelize indefinitely – only to a specific depth, at which time we revert to a serial implementation.  By starting the routine with a maxDepth equal to Environment.ProcessorCount, we can restrict the total amount of parallel operations significantly, but still provide adequate work for each processing core. With this final change, my timings are much better.  On average, I get the following timings: Framework via Array.Sort: 7.3 seconds Serial Quicksort Implementation: 9.3 seconds Naive Parallel Implementation: 14 seconds Parallel Implementation Restricting Depth: 4.7 seconds Finally, we are now faster than the framework’s Array.Sort implementation.

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  • Query to find the data for every month of last year from the given date using mysql?

    - by Salil
    Hi All, I want data for the the last 1 year from the given date For ex:- I have date "2013-06-01" and i want data as follows also data i want is from three table using Group By or something else Month             Amount         Total_Data June 2013     100                5 May 2013       80                  4 -                     100                5 -                     100                5 July 2012       10                  2 I try following query but didn't workout SELECT DATE_FORMAT(rf.period, '%M %Y') as Month , sum(p.amount * ((100-q.amount)/100)) as Amount ,count(distinct q.label_id) as Total Data FROM table1 rf , table2 p , table3 q ,table4 a where rf.period BETWEEN '2013-06-01' AND '2013-06-01' and q.royalty_period BETWEEN '2013-06-01' AND '2013-06-01' and a.id = q.album_id and p.file_id = rf.id and p.upc = a.upc and p.local_revenue is not null GROUP BY Month Thanks in Advance, Salil Gaikwad

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  • How to determine if an item is the last one in a WPF ItemTemplate?

    - by Mike
    Hi everyone, I have some XAML <ItemsControl Name="mItemsControl"> <ItemsControl.ItemTemplate> <DataTemplate> <TextBox Text="{Binding Mode=OneWay}" KeyUp="TextBox_KeyUp"/> </DataTemplate> </ItemsControl.ItemTemplate> </ItemsControl> that's bound to a simple ObservableCollection private ObservableCollection<string> mCollection = new ObservableCollection<string>(); public MainWindow() { InitializeComponent(); this.mCollection.Add("Test1"); this.mCollection.Add("Test2"); this.mItemsControl.ItemsSource = this.mCollection; } Upon hitting the enter key in the last TextBox, I want another TextBox to appear. I have code that does it, but there's a gap: private void TextBox_KeyUp(object sender, KeyEventArgs e) { if (e.Key != Key.Enter) { return; } TextBox textbox = (TextBox)sender; if (IsTextBoxTheLastOneInTheTemplate(textbox)) { this.mCollection.Add("A new textbox appears!"); } } The function IsTextBoxTheLastOneInTheTemplate() is something that I need, but can't figure out how to write. How would I go about writing it? I've considered using ItemsControl.ItemContainerGenerator, but can't put all the pieces together. Thanks! -Mike

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  • how to place last div into right top corner of parent div? (css)

    - by Radek
    can I somehow using css place the block2 in right top corner of block1? Note that block2 must be the (very) last inside html code of block1 or it could be placed after block1. I cannot make it the first element in block1 <html> <head> <style type="text/css"> .block1 {color:red;width:100px;border:1px solid green;} .block2 {color:blue;width:70px;border:2px solid black;position:relative;} </style> </head> <body> <div class='block1'> <p>text</p> <p>text2</p> <div class='block2'>block2</DIV> </div> </body> </html>

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  • how to place last div into right top corner of partent div? (css)

    - by Radek
    can I somehow using css place the block2 in right top corner of block1? Note that block2 must be the (very) last inside html code of block1 or it could be placed after block1. I cannot make it the first element in block1 <html> <head> <style type="text/css"> .block1 {color:red;width:100px;border:1px solid green;} .block2 {color:blue;width:70px;border:2px solid black;position:relative;} </style> </head> <body> <div class='block1'> <p>text</p> <p>text2</p> <div class='block2'>block2</DIV> </div> </body> </html>

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  • Spring can commit Transaction in finally block with RunTimeException in try block [migrated]

    - by Chance Lai
    The project used Spring + Hibernate Sample code: public void method(){ try{ dao.saveA(entityA); throw RuntimeException; dao.saveB(entityB); }catch(RuntimeException e){ throw e; }finally{ dao.saveC(entityC) } } Finally, just entityC will be saved in database in test. I think saveA, saveB, saveC in the same transaction,they should not be committed. In this case, I want to know why entityC is committed. How does Spring do this in the finally block?

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  • Is there some advantage to filling a stack with nils and interpreting the "top" as the last non-nil value?

    - by dwilbank
    While working on a rubymonk exercise, I am asked to implement a stack with a hard size limit. It should return 'nil' if I try to push too many values, or if I try to pop an empty stack. My solution is below, followed by their solution. Mine passes every test I can give it in my IDE, while it fails rubymonk's test. But that isn't my question. Question is, why did they choose to fill the stack with nils instead of letting it shrink and grow like it does in my version? It just makes their code more complex. Here's my solution: class Stack def initialize(size) @max = size @store = Array.new end def pop empty? ? nil : @store.pop end def push(element) return nil if full? @store.push(element) end def size @store.size end def look @store.last end private def full? @store.size == @max end def empty? @store.size == 0 end end and here is the accepted answer class Stack def initialize(size) @size = size @store = Array.new(@size) @top = -1 end def pop if empty? nil else popped = @store[@top] @store[@top] = nil @top = @top.pred popped end end def push(element) if full? or element.nil? nil else @top = @top.succ @store[@top] = element self end end def size @size end def look @store[@top] end private def full? @top == (@size - 1) end def empty? @top == -1 end end

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  • I want a non admin user to install software. What commands do I need to add to sudoers?

    - by Chance
    I want to edit the /etc/sudoers file so that a non-admin user can install software via the Software Center in Linux Mint 10. The reason for this is that I want a user to have the capability to install programs, but not make any other configuration changes to the system. So far I have the following (some of these may not make sense, I was just trying whatever I thought of) username ALL= /usr/bin/aptitude username ALL= /usr/bin/dpkg username ALL= /usr/local/bin/apt-get username ALL= /usr/lib/linuxmint/mintUpdate/mintUpdate.py username ALL= /usr/bin/software-center username ALL= /usr/bin/synaptic So far, it allows me to do updates without asking for my password, but it will not let me install software without entering an admin password. I am aware of this question, How can I set the Software Center to install software for non-root users?, but this goes the route of modifying the PolicyKit, whereas I'm interested in a sudo solution, because it seems a simpler way to go.

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  • Removing a result from Queryset

    - by Enrico
    Is there a simple way to discard/remove the last result in a queryset without affecting the db? I am trying to paginate results in Django, but don't know the total number of objects for a given query. I was planning on using next/previous or older/newer links, so I only need to know if this is the first and/or last page. First is easy to check. To check for the last page I can compare the number of results with the pagesize or make a second query. The first method fails to detect the last page when the number of results in the last set equals the pagesize (ie 100 records get broken into 10 pages with the last page containing exactly 10 results) and I would like to avoid making a second query. My current thought is that I should fetch pagesize + 1 results from the db. If the queryset length equals 11, I know this is not the last page and I want to discard the last result in the queryset before passing the queryset to the template.

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  • JSF 2 f:ajax lifecycle problem

    - by gerry
    The problem is, that if a property is changed during an f:ajax request and a binded panelGroup should be newly created depending on that changed value, the old value is used. This code will explain the problem. Here is the backingbean TestBean: public String getFirst() { return first; } public void setFirst(String first) { this.first = first; } public String getLast() { return last; } public void setLast(String last) { this.last = last; } public String getName(){ return first+" "+last; } public void setDynamicPanel(HtmlPanelGroup panel){ } public HtmlPanelGroup getDynamicPanel(){ Application app = FacesContext.getCurrentInstance().getApplication(); HtmlPanelGroup component = (HtmlPanelGroup)app.createComponent(HtmlPanelGroup.COMPONENT_TYPE); HtmlOutputLabel label1 = (HtmlOutputLabel)app.createComponent(HtmlOutputLabel.COMPONENT_TYPE); label1.setValue(" --> "+getFirst()+" "+getLast()); component.getChildren().add(label1); return component; } and now the jsf/facelet code: <h:form id="form"> <h:panelGrid columns="1"> <h:inputText id="first" value="#{testBean.first}" /> <h:inputText id="last" value="#{testBean.last}" /> <h:commandButton value="Show"> <f:ajax execute="first last" render="name dyn" /> </h:commandButton> </h:panelGrid> <h:outputText id="name" value="#{testBean.name}" /> <h:panelGroup id="dyn" binding="#{testBean.dynamicPanel}" /> </h:form> After the page was initially loaded the outputText and panelGroup shows both "null" as first and last. But after the button is pressed, the outputText is updated well, but the the panelgroup shows again only "null". This is due to the problem, that the "binded method" dynamicPanel is executed before the update of the first and last properties. how can workaround this behaviour or what is wrong with my code?

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  • Optimus Bumblebee Performance

    - by Chance
    After installing Ubuntu 12.04 and Bumblebee, I tested out the performance of the Nvidia card using Minecraft and a few other games. I've noticed it to be way slower than it was on Windows 7. Is this normal? I'm using the GT 630m. From what I've read online, nobody has said it to be slower than Windows. I'm just really curious because I want to use Linux so much more than Windows, but if I don't get the same performance I feel really picky. The Nvidia card is still faster than my Intel graphics on Ubuntu, but it's not as fast as it was for my on Windows. I get 60 - 80 fps on Minecraft on windows, while I get 28 - 48 fps on Ubuntu. Any Ideas why? Thanks so much!

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  • Does a Postgresql dump create sequences that start with - or after - the last key?

    - by bennylope
    I recently created a SQL dump of a database behind a Django project, and after cleaning the SQL up a little bit was able to restore the DB and all of the data. The problem was the sequences were all mucked up. I tried adding a new user and generated the Python error IntegrityError: duplicate key violates unique constraint. Naturally I figured my SQL dump didn't restart the sequence. But it did: DROP SEQUENCE "auth_user_id_seq" CASCADE; CREATE SEQUENCE "auth_user_id_seq" INCREMENT 1 START 446 MAXVALUE 9223372036854775807 MINVALUE 1 CACHE 1; ALTER TABLE "auth_user_id_seq" OWNER TO "db_user"; I figured out that a repeated attempt at creating a user (or any new row in any table with existing data and such a sequence) allowed for successful object/row creation. That solved the pressing problem. But given that the last user ID in that table was 446 - the same start value in the sequence creation above - it looks like Postgresql was simply trying to start creating rows with that key. Does the SQL dump provide the wrong start key by 1? Or should I invoke some other command to start sequences after the given start ID? Keenly curious.

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  • jQuery: How to reverse sortable('serialize') arrays from last to first?

    - by Binyamin
    The discussion begins http://stackoverflow.com/questions/654535/jquery-what-to-do-with-the-list-that-sortableserialize-returns/2920760#2920760 How to reverse it from last to first, updateList.php?id[]=5&id[]=4&id[]=3&id[]=2&id[]=1&&action=update? <ul> <li id="oreder-5">5</li> <li id="oreder-4">4</li> <li id="oreder-3">3</li> <li id="oreder-2">2</li> <li id="oreder-1">1</li> <ul> My code: $(document).ready(function(){ order=[]; $('#list ul').children('li').each(function(idx, elm) { order.push(elm.id.split('-')[1]) }); $.post('updateList.php', {'order[]': order, action: 'update'}); function slideout(){ setTimeout(function(){ $("#response").slideUp("slow", function () {}); }, 2000); } $("#response").hide(); $(function() { $("#list ul").sortable({ opacity: 0.8, cursor: 'move', update: function() { var order = $(this).sortable("serialize") + '&action=update'; $.post("updateList.php", order, function(theResponse){ $("#response").html(theResponse); $("#response").slideDown('slow'); slideout(); }); }}); }); });

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  • How to make last div stretch to fill screen?

    - by Conor
    I have a site I'm trying to build and I've hit one little snag thats driving me insane. Essentially on pages without enough content to fill the viewport, I want to have the last div (my footer, fill the rest of the viewport, but it's currently being cut off. My html looks like this: <body> <div id="header"> </div> <div id="subNav"> </div> <div id="content"> </div> <div id="footer"> </div> </body> I tried using html, body, footer { height:100%; } but that creates much more space then needed, essentially a full screen length of blank content in the footer. How do I get my footer just to fill teh rest of the screen without adding a scroll bar? Thanks in advance, One Frustrated Coder.

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  • What is the last entry in an unfilled array (C++)?

    - by jwaffe
    I put C++ because I'm just starting in C# and I'm not sure if there's a difference. if you declare an array char arr[10] and fill in values for arr[0] through arr[8], what value will be put in arr[9]? a space ' '? An endline '\n'? '\0'? Or is it nothing at all? I'm asking this because I've always used tactics like this char word[20]; for(count = 0 ; count < 20 ; count++) { cout << word[count]; } to print the entire contents of an array, and I was wondering if I could simplify it (e.g., if the last entry was '\0') by using something like this char word[20]; while(word[count] != '\0') { cout << word[count]; } that way, I wouldn't have to remember how many pieces of data were entered into an array if all the spaces weren't filled up. If you know an even faster way, let me know. I tend to make a bunch of mistakes on arrays.

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  • Get last row of many matrices (ASCII text files) and create a new matrix from these rows

    - by nofunsally
    I have over a thousand matrices (6 x 2000, ASCII files, comma delimited) that I generated from MATLAB. I want to get the last row of each matrix / text file and save them in a new matrix / text file. The text files have crazy names so when I load them I can name them whatever. Right now I would do this to achieve my goal: % A = load('crazyname.txt'); % B = load('crazynameagain.txt'); % C = load('crazynameyetagain.txt'); A = [5 5 5; 5 5 5; 1 1 1]; B = [5 5 5; 5 5 5; 2 2 2]; C = [5 5 5; 5 5 5; 3 3 3]; D(1,:)=A(end,:); D(2,:)=B(end,:); D(3,:)=C(end,:); I will create each command (e.g. load, building D step by step) in Excel by combining text cells to create a command. Is there a better way to do this? Could I load / assign the matrices with a name that would better suit them to be used in a for loop? Or is some other MATLAB command that would facilitate this? Thanks.

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  • When pointing to new DNS servers is there any chance of E-mails being lost if the old E-mail hosting service is still up?

    - by LaserBeak
    I am changing webhosts and will be using the new hosts mail servers instead of the old ones. I have created all the correctly named mailboxes on the new service but have also not yet cut ties with the old webhost. I am expecting that even if the new DNS values which point to the new hosts DNS servers and respective SOA\zone file with the new MX values have not yet propagated and an E-mail is directed at the old hosts mail servers as per the mx records in the SOA\zone records which the old hosting provider holds, the E-mail would still come through to the mailbox that's on the old host providers mail servers. So I am just trying to reaffirm if I got this right and it's essentially impossible for me to loose an E-mail since it will hit either the old hosts mail servers or the new ones ? Also is it possible to configure the same E-mail account to check and collect mail from different mail servers by entering multiple pop3 addresses ? And if I choose to keep the old web hosts mail hosting services as a backup by specifying the mx records for it with a lower priority in the SOA records hosted by the new webhost, is it possible to have any incoming E-mails sent to both servers by the mail daemon so I have two copies? Or is my only option having the primary mail server forward the E-mail somehow to the old mailserver ?

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  • How do I de-duplicate a list of nodes in XSLT - and return the last node encountered?

    - by Broam
    I've seen lots of "de-duplicate this xml" questions but everyone wants the first node or the nodes are identical. I have a bit of a bigger puzzle. I have a list of articles in XML, a relevant snippet is shown: <item><key>Article1</key><stamp>100</stamp></item> <item><key>Article1</key><stamp>130</stamp></item> <item><key>Article2</key><stamp>800</stamp></item> <item><key>Article1</key><stamp>180</stamp></item> <item><key>Article3</key><stamp>900</stamp></item> <item><key>Article3</key><stamp>950</stamp></item> <item><key>Article4</key><stamp>990</stamp></item> <item><key>Article5</key><stamp>999</stamp></item> I'd like a list of nodes where the keys are unique and where the last instance is returned, not the first: Stamp (integer) is always increasing for elements of a particular key. Ideally I'd like "largest stamp" but they're always in order so the shortcut is ok. Desired result: (Order doesn't really matter.) <item><key>Article2</key><stamp>800</stamp></item> <item><key>Article1</key><stamp>180</stamp></item> <item><key>Article3</key><stamp>950</stamp></item> <item><key>Article4</key><stamp>990</stamp></item> <item><key>Article5</key><stamp>999</stamp></item> I'm somewhat confused on how to get this list. Any ideas? I'm using the Saxon processor if it matters.

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  • Why does my Perl regular expression only find the last occurrence?

    - by scharan
    I have the following input to a Perl script and I wish to get the first occurrence of NAME="..." strings in each of the <table>...</table> structures. The entire file is read into a single string and the regex acts on that input. However, the regex always returns the last occurrence of NAME="..." strings. Can anyone explain what is going on and how this can be fixed? Input file: ADSDF <TABLE> NAME="ORDERSAA" line1 line2 NAME="ORDERSA" line3 NAME="ORDERSAB" </TABLE> <TABLE> line1 line2 NAME="ORDERSB" line3 </TABLE> <TABLE> line1 line2 NAME="ORDERSC" line3 </TABLE> <TABLE> line1 line2 NAME="ORDERSD" line3 line3 line3 </TABLE> <TABLE> line1 line2 NAME="QUOTES2" line3 NAME="QUOTES3" NAME="QUOTES4" line3 NAME="QUOTES5" line3 </TABLE> <TABLE> line1 line2 NAME="QUOTES6" NAME="QUOTES7" NAME="QUOTES8" NAME="QUOTES9" line3 line3 </TABLE> <TABLE> NAME="MyName IsKhan" </TABLE> Perl Code starts here: use warnings; use strict; my $nameRegExp = '(<table>((NAME="(.+)")|(.*|\n))*</table>)'; sub extractNames($$){ my ($ifh, $ofh) = @_; my $fullFile; read ($ifh, $fullFile, 1024);#Hardcoded to read just 1024 bytes. while( $fullFile =~ m#$nameRegExp#gi){ print "found: ".$4."\n"; } } sub main(){ if( ($#ARGV + 1 )!= 1){ die("Usage: extractNames infile\n"); } my $infileName = $ARGV[0]; my $outfileName = $ARGV[1]; open my $inFile, "<$infileName" or die("Could not open log file $infileName"); my $outFile; #open my $outFile, ">$outfileName" or die("Could not open log file $outfileName"); extractNames( $inFile, $outFile ); close( $inFile ); #close( $outFile ); } #call main();

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  • When i add a bitmap to an array list the last element is duplicated in previous indexes

    - by saxofone2
    I'm trying to implement a personal way of undo/redo in a finger paint-like app. I have in synthesis three objects: the Main class (named ScorePadActivity), the relative Main Layout (with buttons, menus, etc, as well as a View object where I create my drawings), and a third object named ArrayList where i'm writing the undo/redo code. The problem is, when I press the undo button nothing happens, but if I draw anything again "one-time" and press undo, the screen is updated. If I draw many times, to see any changes happen on screen I have to press the undo button the same number of times I have drawn. Seems like (as in title) when I add a bitmap to the array list the last element is duplicated in previous indexes, and for some strange reason, everytime I press the Undo Button, the system is ok for one time, but starts to duplicate until the next undo. The index increase is verified with a series of System.out.println inserted in code. Now when I draw something on screen, the array list is updated with the code inserted after the invocation of touchup(); method in motionEvent touch_up(); } this.arrayClass.incrementArray(mBitmap); mPath.rewind(); invalidate(); and in ArrayList activity; public void incrementArray(Bitmap mBitmap) { this._mBitmap=mBitmap; _size=undoArray.size(); undoArray.add(_size, _mBitmap); } (All Logs removed for clear reading) The undo button in ScorePadActivity calls the undo method in View activity: Button undobtn= (Button)findViewById(R.id.undo); undobtn.setOnClickListener(new View.OnClickListener() { public void onClick(View v) { mView.undo(); } }); in View activity: public void undo() { this.mBitmap= arrayClass.undo(); mCanvas = new Canvas(mBitmap); mPath.rewind(); invalidate(); } that calls the relative undo method in ArrayList activity: public Bitmap undo() { // TODO Auto-generated method stub _size=undoArray.size(); if (_size>1) { undoArray.remove(_size-1); _size=undoArray.size(); _mBitmap = ((Bitmap) undoArray.get(_size-1)).copy(Bitmap.Config.ARGB_8888,true); } return _mBitmap; } return mBitmap and invalidate: Due to my bad English I have made a scheme to make the problem more clear: I have tried with HashMap, with a simple array, I have tried to change mPath.rewind(); with reset();, new Path(); etc but nothing. Why? Sorry for the complex answer, i want give you a great thanks in advance. Best regards

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  • Extracting the layer transparency into an editable layer mask in Photoshop

    - by last-child
    Is there any simple way to extract the "baked in" transparency in a layer and turn it into a layer mask in Photoshop? To take a simple example: Let's say that I paint a few strokes with a semi-transparent brush, or paste in a .png-file with an alpha channel. The rgb color values and the alpha value for each pixel are now all contained in the layer-image itself. I would like to be able to edit the alpha values as a layer mask, so that the layer image is solid and contains only the RGB values for each pixel. Is this possible, and in that case how? Thanks. EDIT: To clarify - I'm not really after the transparency values in themselves, but in the separation of rgb values and alpha values. That means that the layer must become a solid, opaque image with a mask.

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