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  • Best way to delay access to static web pages until services become available with J2EE

    - by Dean Povey
    I have a J2EE application front-ended by a bunch of GWT pages. When the server is starting up, it is possible that these static pages can be accessed before the services required to implement the GWT RPC calls (database etc) are available. I wondering what the best approach is to prevent a user accessing this static content before these services become available. For the purpose of this exercise, assume that there is an isInitialized() method somewhere. I am happy with either a page displaying an error message or simply refusing the connection.

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  • JLabel wont change color twice

    - by Aly
    Hi, I have the following code: public class Test extends JFrame implements ActionListener{ private static final Color TRANSP_WHITE = new Color(new Float(1), new Float(1), new Float(1), new Float(0.5)); private static final Color TRANSP_RED = new Color(new Float(1), new Float(0), new Float(0), new Float(0.1)); private static final Color[] COLORS = new Color[]{ TRANSP_RED, TRANSP_WHITE}; private int index = 0; private JLabel label; private JButton button; public Test(){ super(); setLayout(new BoxLayout(getContentPane(), BoxLayout.Y_AXIS)); label = new JLabel("hello world"); label.setOpaque(true); label.setBackground(TRANSP_WHITE); getContentPane().add(label); button = new JButton("Click Me"); button.addActionListener(this); getContentPane().add(button); pack(); setVisible(true); } @Override public void actionPerformed(ActionEvent e) { if(e.getSource().equals(button)){ label.setBackground(COLORS[index % (COLORS.length - 1)]); } } public static void main(String[] args) { new Test(); } } When I run it I get the label with the TRANSP_WHITE background and then when I click the button this color changes to TRANSP_RED but when I click it again I see no change in color. Does anyone know why? Thanks

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  • OutOfMemoryError creating a tree recursively?

    - by Alexander Khaos Greenstein
    root = new TreeNode(N); constructTree(N, root); private void constructTree(int N, TreeNode node) { if (N > 0) { node.setLeft(new TreeNode(N-1)); constructTree(N-1, node.getLeft()); } if (N > 1) { node.setMiddle(new TreeNode(N-2)); constructTree(N-2, node.getMiddle()); } if (N > 2) { node.setRight(new TreeNode(N-3)); constructTree(N-3, node.getRight()); } Assume N is the root number, and the three will create a left middle right node of N-1, N-2, N-3. EX: 5 / | \ 4 3 2 /|\ 3 2 1 etc. My GameNode class has the following variables: private int number; private GameNode left, middle, right; Whenever I construct a tree with an integer greater than 28, I get a OutOfMemoryError. Is my recursive method just incredibly inefficient or is this natural? Thanks!

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  • Problem using FormLayout in Swing

    - by Dimitri
    Hi I am using the FormLayout. I just discovered it and it's powerful layout manager. I would like to layout 4 components (outlined, properties, tgraph, library) in 3 columns. I want to layout my library component on top of the outlined component in one column, the graph and the properties component in one column. But it doesn't work. Maybe I miss something. Here is my code : private void layoutComponent() { JPanel panel = new JPanel(); FormLayout layout = new FormLayout( "right:p,10dlu,300dlu,left:max(50dlu;p)", "top:pref,center:p,p"); layout.setRowGroups(new int[][]{{1,3}}); PanelBuilder builder = new PanelBuilder(layout,panel); builder.setDefaultDialogBorder(); CellConstraints constraints = new CellConstraints(); builder.add(library, constraints.xy(1, 1)); builder.add(outline,constraints.xy(1, 3)); builder.add(tgraph,constraints.xy(3, 1)); builder.add(properties,constraints.xy(4, 1)); getContentPane().add(panel);r code here } Can someone help plz. Thx :)

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  • Ordering the results of a Hibernate Criteria query by using information of the child entities of the

    - by pkainulainen
    I have got two entities Person and Book. Only one instance of a specific book is stored to the system (When a book is added, application checks if that book is already found before adding a new row to the database). Relevant source code of the entities is can be found below: @Entity @Table(name="persons") @SequenceGenerator(name="id_sequence", sequenceName="hibernate_sequence") public class Person extends BaseModel { @Id @Column(name = "id") @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "id_sequence") private Long id = null; @ManyToMany(targetEntity=Book.class) @JoinTable(name="persons_books", joinColumns = @JoinColumn( name="person_id"), inverseJoinColumns = @JoinColumn( name="book_id")) private List<Book> ownedBooks = new ArrayList<Book>(); } @Entity @Table(name="books") @SequenceGenerator(name="id_sequence", sequenceName="hibernate_sequence") public class Book extends BaseModel { @Id @Column(name = "id") @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "id_sequence") private Long id = null; @Column(name="name") private String name = null; } My problem is that I want to find persons, which are owning some of the books owned by a specific persons. The returned list of persons should be ordered by using following logic: The person owning most of the same books should be at the first of the list, second person of the the list does not own as many books as the first person, but more than the third person. The code of the method performing this query is added below: @Override public List<Person> searchPersonsWithSimilarBooks(Long[] bookIds) { Criteria similarPersonCriteria = this.getSession().createCriteria(Person.class); similarPersonCriteria.add(Restrictions.in("ownedBooks.id", bookIds)); //How to set the ordering? similarPersonCriteria.addOrder(null); return similarPersonCriteria.list(); } My question is that can this be done by using Hibernate? And if so, how it can be done? I know I could implement a Comparator, but I would prefer using Hibernate to solve this problem.

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  • Ho do I extract a Outlook message nested into another using Apache POI - HSMF?

    - by Jan
    I am using Apache POI - HSMF to extract attachments from Outlooks msg-files. It works fine except for nested messages. If an msg is attached to another msg I am able to get the files. If a message is nested I get the information but I need the file. MAPIMessage msg = new MAPIMessage(fileName) for(AttachmentChunks attachment : msg.getAttachmentFiles()) { if(attachment.attachmentDirectory!=null){ MAPIMessage nestedMsg attachment.attachmentDirectory.getAsEmbededMessage(); // now save nestedMsg as a msg-file } } Is it possible to save the nested message file as a regular msg-file?

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  • Easiest way to get Enum in to Key Value pair.

    - by vijay.shad
    Hi, I have defined my Enums like this. public enum UserType { RESELLER("Reseller"), SERVICE_MANAGER("Manager"), HOST("Host"); private String name; private UserType(String name) { this.name = name; } public String getName() { return name; } } What should be the easiest way to get a key-value pair form the enum values?

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  • Android AudioRecorder object wont read from microphone.

    - by supersk
    I'm trying to build a voip application on a new android device, i use AudioRecorder to read the microphone but I'm getting no valid results, just white noise. This happen only on this new device(other work well) and this is probably because this device has PTT ability. Is there some workaround to avoid using AudioRecoder to receive streaming data from the microphone? Thanks. supersk.

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  • How do I delete orphan entities using hibernate and JPA on a many-to-many relationship?

    - by user368453
    I want to delete orphan entities using hibernate and JPA on a many-to-many relationship but all that I found was this atibute the attribute. org.hibernate.annotations.CascadeType.DELETE_ORPHAN ( i.e. @Cascade(value={org.hibernate.annotations.CascadeType.DELETE_ORPHAN) ), which works only for one-to-many relationships. I want to know if I can delete the orphan ones on my many-to-many relationship. I´d be happy if anyone could help me... Thanks in advance !

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  • AdMob won't load programmatically

    - by scottbot95
    I'm need to include ads into my app and I have a settings option to disable ads. so I need to load the ad in code. I copied the code from google to handle that and when I set ads:loadAdOnCreate to true, it works just fine. But if I set it to false and add the two lines AdView adView = (AdView)this.findViewById(R.id.ad); adView.loadAd(new AdRequest()); The ads stop displaying. If I look at log cat, it shows that it is receiving an ad and trying to display it. However it won't actually display on screen. Help?

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  • Hibernate Query - Get latest versions by timestamp?

    - by Slim
    I have a database that is being used as a sort of version control system. That is, instead of ever updating any rows, I add a new row with the same information. Each row also contains a version column that is a date timestamp, so the only difference is the new row will have a more recent timestamp. What I'm having trouble with is writing an efficient hibernate query to return the latest version of these rows. For the sake of example, these are rows in a table called Product, the timestamped column is version. There are multiple versions of multiple products in the table. So there may be multiple versions (rows) of ProductA, multiple versions of ProductB, etc. And I would like to grab the latest version of each. Can I do this in just a single hibernate query? session.createQuery("select product from Product product where...?"); Or would this require some intermediate steps?

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  • JPA and compatibility with persistance providers and databases vendors

    - by Kartoch
    JPA promises to be vendor neutral for persistence and database. But I already know than some persistence frameworks like hibernate are not perfect (character encoding, null comparison) and you need to adapt your schema for each database. Because there is two layers (the persistence framework and database), I would imagine they're some work to use some JPA codes... Does anyone has some experiences with multiple support and if yes, what are the tricks and recommendations to avoid such incompatibilities ?

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  • How to store currently logged on user in DB?

    - by stacker
    Using Seam 2.1.2 and JSF 1.2 I wonder how to store the users login name in the database. In plain JSF I would simply lookup FacesContext.getCurrentInstance().getExternalContext().getRemoteUser();in a backing bean and set the value into a persistent object. How can I achieve that the users name will be stored in the DB?

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  • How to change the jtooltip timers for 1 component.

    - by i30817
    I have a component where i want to display a custom jtooltip. That is easy, just change the getTooltip method. Similar for location and text. However i also want to change the timers. The tooltip should always be displayed if the mouse is over a cellrenderer of the component. If it leaves all of those it should be turned invisible. I know that i can use TooltipManager to control the times globally. But the best solution is probably to just shortcircut that and display the tooltip myself with a mouselistener. However when i tried to do that (unregister the component in TooltipManager and setting the tooltip visible, with text and in the correct position, in a mouse listener) the tooltip never showed at all. What am i doing wrong?

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  • disable the Home Button and Back Button"

    - by michael
    i want way to disable the Home Button & Back Button when click on checkbox in application , my application on version 4.2.2 i have code but not work when click on checkbox work stop to application public void HardButtonOnClick(View v) { boolean checked1 = ((CheckBox) v).isChecked(); if(checked1) { SQLiteDatabase db; db = openOrCreateDatabase("Saftey.db", SQLiteDatabase.CREATE_IF_NECESSARY, null); db.setVersion(1); db.setLocale(Locale.getDefault()); db.setLockingEnabled(true); ContentValues values = new ContentValues(); values.put("hardBtn", "YES"); db.update("Setting", values, "id = ?", new String[] { "1" }); Toast.makeText(this, "Hard Button Locked", Toast.LENGTH_LONG).show(); //SharedPreferences pref = getSharedPreferences("pref",0); //SharedPreferences.Editor edit = pref.edit(); //edit.putString("hard","yes"); //edit.commit(); /* String Lock="yes" ; Bundle bundle = new Bundle(); bundle.putString("key", Lock); Intent a = new Intent(Change_setting.this, ChildMode.class); a.putExtras(bundle); startActivity(a);*/ super.onAttachedToWindow(); this.getWindow().setType(WindowManager.LayoutParams.TYPE_KEYGUARD); isLock = true; } else { SQLiteDatabase db; db = openOrCreateDatabase("Saftey.db", SQLiteDatabase.CREATE_IF_NECESSARY, null); db.setVersion(1); db.setLocale(Locale.getDefault()); db.setLockingEnabled(true); ContentValues values = new ContentValues(); values.put("hardBtn", "NO"); db.update("Setting", values, "id = ?", new String[] { "1" }); //SharedPreferences pref = getSharedPreferences("pref",0); //SharedPreferences.Editor edit = pref.edit(); //edit.putString("hard","no"); //edit.commit(); Toast.makeText(this, "Hard Button Un-Locked", Toast.LENGTH_LONG).show(); isLock = false; } } how can work it??

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  • which design choose? - pros and cons

    - by Guarava Makanili
    Which of these 3 approches would choose and why? // This is the one I would choose class Car { } class FeeCalculator { public double calculateFee(Car car) { return 0; } } // in that case the problem might be when we use ORM framework and we try to invoke save with parameter Car class Car { private FeeCalculator calculator; public double calculateFee() { return calculator.calculateFee(this); } } class FeeCalculator { public double calculateFee(Car car) { return 0; } } // in that case the problem mentioned above is solved, but I don't like this design class Car { public double calculateFee(FeeCalculator calculator) { return calculator.calculateFee(this); } } class FeeCalculator { public double calculateFee(Car car) { return 0; } }

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  • Line up swing components by edges

    - by rasen58
    Is it possible to line up swing components? The components are in separate panels which both use flow layout. These two panels are in another panel which is using a grid layout. As you can see there is a subtle difference and I find it annoying. I know that all of the jlabels [the rectangles in blue/purple all have the same size, so i think it might be because of the '+' and '*', but I'm not sure because the left sides of the first two boxes aren't lined up. the panels JPanel panel2 = new JPanel(new GridLayout(4, 1)); JPanel panel2a = new JPanel(new FlowLayout()); JPanel panel2b = new JPanel(new FlowLayout()); the first two rectangles (purple) add1 = new JLabel("", JLabel.CENTER); add1.setTransferHandler(new TransferHandler("text")); add1.setBorder(b2); add2 = new JLabel("", JLabel.CENTER); add2.setTransferHandler(new TransferHandler("text")); add2.setBorder(b2); the two blue rectangles textFieldA = new JTextField(); textFieldA.setHorizontalAlignment(JTextField.CENTER); textFieldA.setEditable(false); textFieldA.setBorder(new LineBorder(Color.blue)); textFieldM = new JTextField(); textFieldM.setHorizontalAlignment(JTextField.CENTER); textFieldM.setEditable(false); textFieldM.setBorder(new LineBorder(Color.blue)); the + and * opA = new JLabel("+", JLabel.CENTER); opS = new JLabel("*", JLabel.CENTER); Showing that the rectangles are the same size Dimension d = card1.getPreferredSize(); int width = d.width + 100; int height = d.height + 50; add1.setPreferredSize(new Dimension(width, height)); add2.setPreferredSize(new Dimension(width, height)); mult1.setPreferredSize(new Dimension(width, height)); mult2.setPreferredSize(new Dimension(width, height)); textFieldA.setPreferredSize(new Dimension(width, height)); textFieldM.setPreferredSize(new Dimension(width, height)); Adding to the panels panel2a.add(add1); panel2a.add(opA); panel2a.add(add2); panel2a.add(enterA); panel2a.add(textFieldA); panel2c.add(mult1); panel2c.add(opM); panel2c.add(mult2); panel2c.add(enterM); panel2c.add(textFieldM); panel2.add(panel2a); panel2.add(panel2c);

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  • Linked List Inserting strings in alphabetical order

    - by user69514
    I have a linked list where each node contains a string and a count. my insert method needs to inset a new node in alphabetical order based on the string. if there is a node with the same string, then i increment the count. the problem is that my method is not inserting in alphabetical order public Node findIsertionPoint(Node head, Node node){ if( head == null) return null; Node curr = head; while( curr != null){ if( curr.getValue().compareTo(node.getValue()) == 0) return curr; else if( curr.getNext() == null || curr.getNext().getValue().compareTo(node.getValue()) > 0) return curr; else curr = curr.getNext(); } return null; } public void insert(Node node){ Node newNode = node; Node insertPoint = this.findIsertionPoint(this.head, node); if( insertPoint == null) this.head = newNode; else{ if( insertPoint.getValue().compareTo(node.getValue()) == 0) insertPoint.getItem().incrementCount(); else{ newNode.setNext(insertPoint.getNext()); insertPoint.setNext(newNode); } } count++; }

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