Search Results

Search found 46894 results on 1876 pages for 'java native interface'.

Page 1024/1876 | < Previous Page | 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031  | Next Page >

  • Query in data population using select in jsp

    - by sarah
    I am populating data using <select name="test"> <option value='<%=session.getAttribute("tList")%>'><%=session.getAttribute("tList") %></option> </select> but the values are getting display in a single row in the combo box not row wise,where i am going wrong ?

    Read the article

  • Word filter that groups words?

    - by Legend
    Is there any library that achieves the following: Convert Microsoft Windows 98 Microsoft Windows XP Windows 7 Windows Ultimate Desktop Windows to Windows 4 The complicated part here is to recognize that "Desktop Windows" is an anomaly here and not count it. If nothing is added before the word "Windows", perhaps it can be counted but if there is something else and the suffix does not match any popular suffix, it can still be counted. Maybe I am a little vague here but perhaps someone could have an idea about what I am talking about here. Any suggestions?

    Read the article

  • Update View at runtime in Android

    - by seretur
    The example is pretty straightforward: i want to let the user know about what the app is doing by just showing a text (canvas.drawText()). Then, my first message appears, but not the other ones. I mean, i have a "setText" method but it doesn't updates. onCreate(Bundle bundle) { super.onCreate(bundle); setContentView(splash); // splash is the view class loadResources(); splash.setText("this"); boundWebService(); splash.setText("that"): etc(); splash.setText("so on"); } The view's text drawing works by doing just a drawText in onDraw();, so setText changes the text but doesn't show it. Someone recommended me replacing the view with a SurfaceView, but it would be alot of trouble for just a couple of updates, SO... how the heck can i update the view dinamically at runtime? It should be quite simple, just showing a text for say 2 seconds and then the main thread doing his stuff and then updating the text... Thanks! Update: I tried implementing handler.onPost(), but is the same story all over again. Let me put you the code: package coda.tvt; import android.app.Activity; import android.graphics.Canvas; import android.graphics.Paint; import android.os.Bundle; import android.view.View; import android.widget.TextView; import android.widget.Toast; public class ThreadViewTestActivity extends Activity { Thread t; Splash splash; /** Called when the activity is first created. */ @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); splash = new Splash(this); t = new Thread(splash); t.start(); splash.setTextow("OA"); try { Thread.sleep(4000); } catch (InterruptedException e) { } splash.setTextow("LALA"); } } And: public class Splash implements Runnable { Activity activity; final Handler myHandler = new Handler(); public Splash(Activity activity) { this.activity=activity; } @Override public void run() { // TODO Auto-generated method stub } public synchronized void setTextow(final String textow) { // Wrap DownloadTask into another Runnable to track the statistics myHandler.post(new Runnable() { @Override public void run() { TextView t = (TextView)activity.findViewById(R.id.testo); t.setText(textow); t.invalidate(); } }); } } Although splash is in other thread, i put a sleep on the main thread, i use the handler to manage UI and everything, it doesn't changes a thing, it only shows the last update.

    Read the article

  • new JDK 7 features

    - by xdevel2000
    I wish to test the new features that will came with the next JDK like project coin, project lambda etc. but the last JDK 7 to download will not have any already implemented! From which build can I test them? I think it's incredible that, now in may 2010 at few months to the official final release (november 2010????) for we developers there is no possibility to test any of this features!!

    Read the article

  • How do I remove an old JPanel and add a new one?

    - by Roman
    I would like to remove an old JPanel from the Window (JFrame) and add a new one. How should I do it? I tried the following: public static void showGUI() { JFrame frame = new JFrame("Colored Trails"); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); frame.add(partnerSelectionPanel); frame.setSize(600,400); frame.setVisible(true); } private static void updateGUI(final int i, final JLabel label, final JPanel partnerSelectionPanel) { SwingUtilities.invokeLater( new Runnable() { public void run() { label.setText(i + " seconds left."); } partnerSelectionPanel.setVisible(false); \\ <------------ } ); } So, my code update the "old" JPanel and them it makes the whole JPanel invisible. It was the idea. But it does not work. The compiler complains about the line indicated with "<------------". It writes: <identifier> expected, illegal start of type.

    Read the article

  • Maven + Tomcat acceleration

    - by Bar
    I am writing a web application with Maven in the Eclipse IDE, and use Tomcat servlet container. So, I run Maven like this: mvn clean compile. It is reasonable that after this oepration I must re-run Tomcat so it can reinitialize the context (Sysdeo Tomcat launcher helps a lot). The problem is Maven execution and subsequebt Tomcat re-running takes noticable amount of time (like 10+ seconds for Maven and 20+ sec. for Tomcat, because of logging, Hibernate mappings, etc.) every time I do it. Is there any automated and more faster solution for these two operatioins? As I see it, a way better solution can be moving re-compiled classes only to the target dir.

    Read the article

  • More than one JPanel in a Frame / having a brackground Image and another Layer with Components on the top

    - by user1905203
    I've got a JFrame with a JPanel in which there is a JLabel with an ImageIcon(). Everything's working perfectly, problem is i now want to add another JPanel with all the other stuff like buttons and so on to the JFrame. But it still shows the background Image on top and nothing with the second JPanel. Can someone help me? Here is an extract of my code: JFrame window = new JFrame("Http Download"); /* * Background Section */ JPanel panel1 = new JPanel(); JLabel lbl1 = new JLabel(); /* * Component Section */ JPanel panel2 = new JPanel(); JLabel lbl2 = new JLabel(); /* * Dimension Section */ Dimension windowSize = new Dimension(800, 600); Dimension screen = Toolkit.getDefaultToolkit().getScreenSize(); public HTTPDownloadGUI() { window.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); panel1.setLayout(null); panel1.setSize(windowSize); panel1.setOpaque(false); panel2.setLayout(null); panel2.setSize(windowSize); panel2.setOpaque(false); lbl1.setSize(windowSize); lbl1.setLocation(0, 0); lbl1.setIcon(new ImageIcon(getClass().getResource("bg1.png"))); panel1.add(lbl1); lbl2.setBounds(0, 0, 100, 100); //lbl2.setIcon(new ImageIcon(getClass().getResource("bg2.png"))); lbl2.setBackground(Color.GREEN); panel2.add(lbl2); panel1.add(panel2); window.add(panel1); int X = (screen.width / 2) - (windowSize.width / 2); int Y = (screen.height / 2) - (windowSize.height / 2); window.setBounds(X,Y , windowSize.width, windowSize.height); window.setVisible(true); }

    Read the article

  • Is this Leftist Tree piece of code from Wikipedia correct?

    - by they changed my name
    Link public Node merge(Node x, Node y) { if(x == null) return y; if(y == null) return x; // if this was a max height biased leftist tree, then the // next line would be: if(x.element < y.element) if(x.element.compareTo(y.element) > 0) { // x.element > y.element Node temp = x; x = y; y = temp; } x.rightChild = merge(x.rightChild, y); if(x.leftChild == null) { // left child doesn't exist, so move right child to the left side x.leftChild = x.rightChild; x.rightChild = null; x.s = 1; } else { // left child does exist, so compare s-values if(x.leftChild.s < x.rightChild.s) { Node temp = x.leftChild; x.leftChild = x.rightChild; x.rightChild = temp; } // since we know the right child has the lower s-value, we can just // add one to its s-value x.s = x.rightChild.s + 1; } return x; } What makes me ask this question is: if(x.element.compareTo(y.element) > 0) { // x.element > y.element Node temp = x; x = y; y = temp; } Isn't that just not gonna work, since the references are only switched inside the method?

    Read the article

  • Can I have a set containing identical elements?

    - by Roman
    It is convenient for me to use a set. I like how I can "add" ("remove") an element to (from) the set. It is also convenient to check if a given element is in the set. The only problem, I found out that I cannot add a new element to a set if the set has already such an element. Is it possible to have "sets" which can contain several identical elements.

    Read the article

  • Creating a assertClass() method in JUnit

    - by Mike
    Hi, I'm creating a test platform for a protocol project based on Apache MINA. In MINA when you receive packets the messageReceived() method gets an Object. Ideally I'd like to use a JUnit method assertClass(), however it doesn't exist. I'm playing around trying to work out what is the closest I can get. I'm trying to find something similar to instanceof. Currently I have: public void assertClass(String msg, Class expected, Object given) { if(!expected.isInstance(given)) Assert.fail(msg); } To call this: assertClass("Packet type is correct", SomePacket.class, receivedPacket); This works without issue, however in experimenting and playing with this my interest was peaked by the instanceof operator. if (receivedPacket instanceof SomePacket) { .. } How is instanceof able to use SomePacket to reference the object at hand? It's not an instance of an object, its not a class, what is it?! Once establishing what type SomePacket is at that point is it possible to extend my assertClass() to not have to include the SomePacket.class argument, instead favouring SomePacket?

    Read the article

  • Need an efficient algorithm solve this kind of complex structure

    - by Rizvan
    Problem Statement is : Given 2 Dimensional array, print output for example If 4 rows and 6 columns, output would be: 1 2 3 4 5 6 16 17 18 19 20 7 15 24 23 22 21 8 14 13 12 11 10 9 I tried it is looking like square within square but when I attempted this problem, I put so many while and if loops but didn't got exact answer. If row and columns increases how to handle it? This is not homework. I was learning solving complex structure so I need to understand it by some guidance.

    Read the article

  • Switching application-wide theme programmatically?

    - by Cheezmeister
    EDIT: Related question here: Multi theme support in android app I'm attempting to get a user-chosen theme and feel like I'm frustratingly close. Defining the theme in AndroidManifest.xml works as it should, but (as best I can tell) can't change based on app preferences: <application android:theme="@style/theme_sunshine" android:icon="@drawable/icon" android:label="@string/app_name"> Alternatively, setting it dynamically in each activity also works: someChosenTheme = PreferenceManager.getDefaultSharedPreferences(this).getString("themePreference", "theme_twilight"); setTheme(someOtherChosenTheme); But that seems messy, and I'd rather set the theme for the entire app in one place. My first thought was to grab the application context as soon as my main activity launches and do it there: getApplicationContext().setTheme(R.style.theme_dummy); As best I can tell[0], this ought to do the trick, but in fact it's not doing anything--the entire app has the default Android style. Is the above valid, and if so, might I be doing something else dumb? I'm working in API level 3 if that matters. Prods in the right direction greatly appreciated! [0] http://developer.android.com/reference/android/content/Context.html#setTheme%28int%29 http://developer.android.com/reference/android/content/Context.html#getApplicationContext%28%29

    Read the article

  • Hibernate Performance Best Practice?

    - by user829237
    Im writing a Web application using Hibernate 3. So, after a while i noticed that something was slow. So i tested hibernate profiler and found that hibernate will make unreasonably many db-calls for simple operation. The reason is ofcourse that i load an Object (this object has several "parents") and these "parents" have other "parents". So basicly hibernate loads them all, even though i just need the basic object. Ok, so i looked into lazy-loading. Which lead me into the Lazyloading-exception, because i have a MVC webapp. So now i'm a bit confused as to what is my best approach to this. Basicly all I need is to update a single field on an object. I already have the object-key. Should I: 1. Dig into Lazy-loading. And then rewrite my app for a open-session-view? 2. Dig into lazy-loading. And then rewrite my dao's to be more specific. E.g. writing DAO-methods that will return objects instanciated with only whats necessary for each use-case? Could be a lot of extra methods... 3. Scratch hibernate and do it myself? 4. Cant really think of other solutions right now. Any suggestions? What is the best practice?

    Read the article

  • Sort a list whit element still in first position

    - by Mercer
    Hello, i have a String list List<String> listString = new ArrayList<String>(); listString.add("faq"); listString.add("general"); listString.add("contact"); I do some processing on the list and i want to sort this list but I want general is still in first position Thx ;)

    Read the article

  • Regular Expression Program

    - by david robers
    Hi I have the following text: SMWABCCA ABCCAEZZRHM NABCCAYJG XABCCA ABCCADK ABCCASKIYRH ABCCAKY PQABCCAK ABCCAKQ This method takes a regex in out by the user and SHOULD print out the Strings it applies to but seems to print out something completely different: private void matchIt(String regex) { Pattern p = Pattern.compile(regex); Matcher m = null; boolean found = false; for(int i = 0; i < data.length; i++){ m = p.matcher(data[i]); if(m.find()){ out.println(data[i]); found = true; } } if(!found){ out.println("Pattern Not Found"); } } When inputting "[C]" It outputs: SMWABCCA ABCCAEZZRHM NABCCAYJG XABCCA ABCCADK ABCCASKIYRH ABCCAKY PQABCCAK ABCCAKQ Any ideas why? I think I'm using m.find() improperly...

    Read the article

  • Connecting to an RMI server that sits behind a firewall?

    - by MalcomTucker
    I know my RMI app works correctly - it works fine when the server is on localhost and inside the LAN but when connecting to an external RMI server it fails when trying to make stub calls So the server is bound to localhost (an internal IP - 192.168.1.73) but the client is specifying an external IP (45.4.234.56) - which then gets forwarded to the internal server. How do you resolve this problem? thanks

    Read the article

  • hibernate pagination mechanism

    - by haicnpmk44
    I am trying to use Hibernate pagination for my query (PostgreSQL ) i set setFirstResult(0), setMaxResults(20) for my sql query. My code like below: Session session = getSessionFactory().getCurrentSession(); session.beginTransaction(); Query query = session.createQuery("select id , customer_name , address from tbl_customers "); query.setFirstResult(0); query.setMaxResults(20); List<T> entities = query.list(); session.getTransaction().commit(); but when viewing SQL hibernate log, i still see full sql query: Hibernate: select customer0_.id as id9_, customer0_.customer_name as dst2_9_, customer0_.addres as dst3_9_ from tbl_customers customer0_ Why there is no LIMIT OFFSET in query of Hibernate pagination SQL log? Does anyone know about Hibernate pagination mechanism? I guess that Hibernate will select all data, put data into Resultset, and then paging in Resultset, right?

    Read the article

  • Format String become 0001, 0010 etc

    - by trycatch4j
    Hi all.., I have number : 1, 2, 3, 4, 10 But I wanna print that number 0001 0002 0003 0004 0010 I have search in google, the keyword is number format. but I've got nothing, I just get, frmat decimal such ass 1,000,000.00. hope you can suggest me a reference or give me some problem solving. Thanks,

    Read the article

< Previous Page | 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031  | Next Page >