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  • No response in Eclipse: File ->Import->Existing Projects into Workspace

    - by Hula
    I'm trying to import one of the GWT samples into Eclipse by following the instructions below. But when I browse to the directory containing the "Hello" sample and uncheck "Copy projects into workspace", the Finish button is grayed out, preventing me from completing the import. Any ideas why? -- Option A: Import your project into Eclipse (recommended) -- If you use Eclipse, you can simply import the generated project into Eclipse. We've tested against Eclipse 3.3 and 3.4. Later versions will likely also work, earlier versions may not. In Eclipse, go to the File menu and choose: File - Import... - Existing Projects into Workspace Browse to the directory containing this file, select "Hello". Be sure to uncheck "Copy projects into workspace" if it is checked. Click Finish.

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  • How can you remove a Criterion from a criteria?

    - by ChuckM
    Hello, For instance if I do something like: Criteria c = session.createCriteria(Book.class) .add(Expression.ge("release",reDate); .add(Expression.ge("price",price); .addOrder( Order.asc("date") ) .setFirstResult(0) .setMaxResults(10); c.list(); How can I use the same criteria instance, but remove (for example) the second criterion? I'm trying to build a dynamic query in which I'd like to let the user remove a filter, without the backend having to reconstruct the criteria from scratch. Thank you

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  • How to manage a BorderLayout with generic JPanel()

    - by Nick G.
    Im not sure how to reference to JPanel when it was declared like this, I had someone else help me on JPanels and this is the code he used: final JFrame frame = new JFrame("CIT Test Program"); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); frame.setPreferredSize(new Dimension(350, 250)); frame.add(new JPanel() {{ Not sure how to reference to JPanel to use BorderLayout. How would I go about doing this?

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  • comma in regex in String.replaceAll() method?

    - by kknight
    My code tries to replace "," with "/" in a string. Should I escape "," in the regex string? Both of the two code snippets generated the same results, so I am confused. Code snippet 1: String test = "a,bc,def"; System.out.println(test.replaceAll("\\,", "/")); Code snippet 2: String test = "a,bc,def"; System.out.println(test.replaceAll(",", "/")); Should I use "," or "\,"? Which is safer? Thanks.

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  • jboss web services to tranfer files

    - by mengmenger
    I dont have any experience for web services. please give me some suggestion about the task below. the task is: users will send a txt file (size should less than 20K ) from a .NET application, I need to write a web services which runs by jboss 5.x to read this file and edit this file and send the file back to .NET UI to display the edited version. question is that if the txt file is just test string or binary string, are there any restriction of the string length? if it's binary string, can I need to use BinaryReader class to read it? or not need special reader to read it? (this could be a dumn question :P) what if the .NET application can save the file on either the .NET application server or some shared server location, send a download URL to web serivces, can web services download it and read it? JBoss will be run on Linux sever. After edit the file, how do I send it back? Thanks for your help!

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  • Strange "cache" effect between client and server

    - by mark
    I use a Socket-based connection between Client and server with ObjectOutputStream. The objects serialized and exchanged have this structure: public class RichiestaSalvataggioArticolo implements Serializable { private ArticoloDati articolo; public RichiestaSalvataggioArticolo(ArticoloDati articolo) { this.articolo = articolo; } @Override public void ricevi(GestoreRichieste gestore) throws Exception { gestore.interpreta(this); } public ArticoloDati getArticolo() { return articolo; } } the issue is that when I try to exchange messages between C/S with incapsulated content very similar (ArticoloDati whom differ only in 2 fields out of 10), the Client sends an ArticoloDati, but the Server receives the previous one. Does the ObjectOutputStream implement some kind of cache or memory between the calls, that fails to recognize that my 2 objects are different because they are very similar?

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  • How to debug ConcurrentModificationException?

    - by Dani
    I encountered ConcurrentModificationException and by looking at it I can't see the reason why it's happening; the area throwing the exception and all the places modifying the collection are surrounded by synchronized (this.locks.get(id)) { ... } // locks is a HashMap<String, Object>; I tried to catch the the pesky thread but all I could nail (by setting a breakpoint in the exception) is that the throwing thread owns the monitor while the other thread (there are two threads in the program) sleeps. How should I proceed? What do you usually do when you encounter similar threading issues?

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  • How to create live stream audio for web-sites ???

    - by Kathir
    Hi All, We are storing sound from mic to pc via sound forge. We would like to broadcast the sound which comes from the mic to the pc as live streaming audio. Basically a person speaks in a mic, we like to give it as live stream audio. The web-site is hosted on yahoo server. Can you please let me know in what are the ways we can achieve this? Thanks, Kathir

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  • J2ME development and native API

    - by Attilah
    Is it possible to write a mobile application with J2ME and whenever we want to implement a functionality not offered by J2ME call native mobile API ? (kind of like what is done with .NET, whenever you need something not provided, you just call the Win32 API from the .NET platform).

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  • JVM with no garbage collection

    - by HH
    I've read in many threads that it is impossible to turn off garbage collection on Sun's JVM. However, for the purpose of our research project we need this feature. Can anybody recommend a JVM implementation which does not have garbage collection or which allows turning it off? Thank you.

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  • rule based file parsing

    - by user359490
    I need to parse a file line by line on given rules. Here is a requirement. file can have multiple lines with different data.. 01200344545143554145556524341232131 1120034454514355414555652434123213101200344545143554145556524341232131 2120034454514 and rules can be like this. if byte[0,1] == "0" then extract this line to /tmp/record0.dat if byte[0,1] == "1" then extract this line to /tmp/record1.dat if byte[0,1] == "2" then extract this line to /tmp/record2.dat I am looking for any language which can do this in a fast manner with a very long file size like 2 GB. Appreciate all the help in advance. Thanks

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  • Extract elements from list based on object property type

    - by Dustin Digmann
    Often, I have a list of objects. Each object has properties. I want to extract a subset of the list where a specific property has a predefined value. Example: I have a list of User objects. A User has a homeTown. I want to extract all users from my list with "Springfield" as their homeTown. I normally see this accomplished as follows: List users = getTheUsers(); List returnList = new ArrayList(); for (User user: users) { if ("springfield".equalsIgnoreCase(user.getHomeTown()) returnList.add(user); } I am not particularly satisfied with this solution. Yes, it works, but it seems so slow. There must be a non-linear solution. Suggestions?

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  • Commons VFS and IBM MVS System

    - by Liming
    Hello All, I'm using Apache Commons VFS / SFTP, we are trying to download files from the IBM MVS system. The download part is all good, however, we can not open up the zipped files after downloading. Seems like the zip file was compressed using a different algorithm or something Anyone has any pointers? *Note, the same function works fine if we connect to a regular unix/linux SFTP server. Below is an example of what we did String defaultHost = "[my sftp ip address]"; String host = defaultHost; String defaultRemotePath = "//__root.dir1.dir2."; String remotePath = defaultRemotePath; String user = "test"; String password = "test"; String remoteFileName = "Blah.ZIP.BLAH"; log.info("FtpPojo() begin instantiation"); FileObject localFileObject = fsManager.resolveFile("C:/Work/Blah.ZIP.BLAH"); log.debug("local file name is :"+localFileObject.getName().getBaseName()); log.debug("FtpPojo() instantiated and fsManager created"); String uri = createSftpUri(host, user, password) + ":322"+remotePath+remoteFileName; remoteRepo = fsManager.resolveFile(uri, fsOptions); remoteRepo.copyFrom(localFileObject, Selectors.SELECT_ALL);

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  • After a few same hql query the application freezes

    - by Oktay
    I am calling below function with the same batchNumber and it is working without problem 15 times and takes the records fromm database without problem but at 16. time the application freezes when the query.list() row is called. It just loses debug focus and not give any exception. This problem probably is not about the hql because I've seen this problem before and I used criteria instead of hql and I got pass this problem. But for this when I use "group by" in criteria(setrojection....) it doesn't return the result as hibernate model(object) just returns a list. But I need the results as model. Note: about 15 times it is just for test. This is a web aplication and user may click the button many times that calls this funtion to see the taken records from database. public List<SiteAddressModel> getSitesByBatch(String batchNumber) { try{ List<SiteAddressModel> siteList; MigrationPlanDao migrationPlanDao = ServiceFactory.getO2SiteService().getMigrationPlanDao(); Query query = this.getSession().createQuery("from " + persistentClass.getName() + " where " + "siteType =:" + "type and siteName in " + "(select distinct exchange from " + migrationPlanDao.getPersistentClass().getName() + " where migrationBatchNumber =:" + "batchNumber" + ")" ); query.setString("batchNumber", batchNumber); query.setString("type", "LLU/ASN"); System.out.println("before query"); siteList = query.list(); System.out.println("after query"); return siteList; }catch (Exception e) { e.printStackTrace(); } Hibernate version 3.2.0.ga

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  • mvc design in a card game

    - by Hong
    I'm trying to make a card game. some classes I have are: CardModel, CardView; DeckModel, DeckView. The deck model has a list of card model, According to MVC, if I want to send a card to a deck, I can add the card model to the deck model, and the card view will be added to the deck view by a event handler. So I have a addCard(CardModel m) in the DeckModel class, but if I want to send a event to add the card view of that model to the deck view, I only know let the model has a reference to view. So the question is: If the card model and deck model have to have a reference to their view classes to do it? If not, how to do it better? Update, the code: public class DeckModel { private ArrayList<CardModel> cards; private ArrayList<EventHandler> actionEventHandlerList; public void addCard(CardModel card){ cards.add(card); //processEvent(event x); //must I pass a event that contain card view here? } CardModel getCards(int index){ return cards.get(index); } public synchronized void addEventHandler(EventHandler l){ if(actionEventHandlerList == null) actionEventHandlerList = new ArrayList<EventHandler>(); if(!actionEventHandlerList.contains(l)) actionEventHandlerList.add(l); } public synchronized void removeEventHandler(EventHandler l){ if(actionEventHandlerList!= null && actionEventHandlerList.contains(l)) actionEventHandlerList.remove(l); } private void processEvent(Event e){ ArrayList list; synchronized(this){ if(actionEventHandlerList!= null) list = (ArrayList)actionEventHandlerList.clone(); else return; } for(int i=0; i<actionEventHandlerList.size(); ++i){ actionEventHandlerList.get(i).handle(e); } } }

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  • uploading app problem in developers site

    - by Siva K
    hi i have posted one of my app in market.android.com when i am trying to post second app it shows "You have another application on Market with the same package name. Go to that other application, and click upgrade" I dont want to upgrade it but i want my second app to be posted. When i tried to change the package name it showed lots of errors, so i decided to create the app once again in a new package and project name, it seems to be a very lengthy process.... pls help me to solve the issue bcoz i have created all my app in same package name unknowingly.......

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