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  • How to create a datastore.Text object out of an array of dynamically created Strings?

    - by Adrogans
    I am creating a Google App Engine server for a project where I receive a large quantity of data via an HTTP POST request. The data is separated into lines, with 200 characters per line. The number of lines can go into the hundreds, so 10's of thousands of characters total. What I want to do is concatenate all of those lines into a single Text object, since Strings have a maximum length of 500 characters but the Text object can be as large as 1MB. Here is what I thought of so far: public void doPost(HttpServletRequest req, HttpServletResponse resp) { ... String[] audioSampleData = new String[numberOfLines]; for (int i = 0; i < numberOfLines; i++) { audioSampleData[i] = req.getReader().readLine(); } com.google.appengine.api.datastore.Text textAudioSampleData = new Text(audioSampleData[0] + audioSampleData[1] + ...); ... } But as you can see, I don't know how to do this without knowing the number of lines before-hand. Is there a way for me to iterate through the String indexes within the Text constructor? I can't seem to find anything on that. Of note is that the Text object can't be modified after being created, and it must have a String as parameter for the constructor. (Documentation here) Is there any way to this? I need all of the data in the String array in one Text object. Many Thanks!

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  • Netbeans configuration problem

    - by Yatendra Goel
    I am using Netbeans 6.8 The problem is that the projects explorer (that displays all the projects and their contents) displays each package as a node. For instance, if there is package hierarchy like this; com.mycompany.myproject.package1.package1.1 then it displays 5 nodes for the five packages which is very disturbing while development. Is there any way by which I can configure it(Netbeans) so that it groups all the subpackages of a package under one node and displays the subpackages only when I expand the package node?

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  • Noob question about hibernate criteria

    - by Dimitri
    Hello, I have a class called User which has 2 properties : login/password. I am trying to authenticate a user in my application using hibernate criteria but my request doesn't work. [EDIT] The returned value is NULL. I have two users in my database for testing. Here is my code : @Override public User authenticate(String login, String password) throws NullPointerException { Session session = this.getSession(); User user = (User) session .createCriteria(User.class) .add( Restrictions.and( Property.forName("login").eq(login), Property.forName("password").eq(password) )).uniqueResult(); if (user == null){ throw new NullPointerException("User not found"); } return user; } Can someone tells me what is wrong with my code? Happy new Year 2011 !!

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  • Creating New Object of Other Class dynamicly ?

    - by Meko
    I am trying to create new object of other class in a for loop. like for(int i =0;i<10;i++){ Computer p1=new Computer(10,20); } and when I try anywhere to reach p1.someAction(); it say you must declare p1. But if I declare it on top of program how can I create again in loop? I also try only Computer p1; but it gave exeption ..

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  • Why does eclipse give me errors when i try to run sample application?

    - by ylen
    I dont know why the sample application from the android website gives me 300+ errors when i try to run it in ecplise galileo. The application i am trying is Bluetoothchat it is straight from the sdk sample folder so it shouldn't contain any. I have added android.jar and I do have an emulator. I have tried HelloWorld and it worked..Could someone help me? Thanks

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  • Resultset To List

    - by Gnaniyar Zubair
    I want to convert my Resultset to List in my JSP page. and want to display all the values. This is my query: SELECT userId, userName FROM user; I have executed that using preparedstatement and got the Resultset. But how to convert it as a List and want to display the result like this: userID userName ------------------ 1001 user-X 1006 user-Y 1007 user-Z

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  • 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); }

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  • How to insert into data base using multi threading programming [closed]

    - by user1196650
    I am having a method and that method needs to do the following thing: It has to insert records into a database. No insert is done for the same table again. All inserts are into different tables. I need a multi threading logic which inserts the details into db using different threads. I am using oracle db and driver configuration and remaining stuff are perfect. Please help me with an efficient answer. Can anyone could provide me with a skeleton logic of the program.

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  • GridBagConstraints weightx and weighty values

    - by xdevel2000
    In many books weightx and weighty values are expressed in different ways: some says 0.0 to 1.0 other says 0 to 100 other say until 1000 I'm a lot confused. In the API these variables are double types so I think the first is correct but what does it meaning a value of 0.4 or 0.7? are percentage values, point values? relative of what?

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  • How to calculate the y-pixels of someones weight on a graph? (math+programming question)

    - by RexOnRoids
    I'm not that smart like some of you geniuses. I need some help from a math whiz. My app draws a graph of the users weight over time. I need a surefire way to always get the right pixel position to draw the weight point at for a given weight. For example, say I want to plot the weight 80.0(kg) on the graph when the range of weights is 80.0 to 40.0kg. I want to be able to plug in the weight (given I know the highest and lowest weights in the range also) and get the pixel result 400(y) (for the top of the graph). The graph is 300 pixels high (starts at 100 and ends at 400). The highest weight 80kg would be plot at 400 while the lowest weight 40kg would be plot at 100. And the intermediate weights should be plotted appropriately. I tried this but it does not work: -(float)weightToPixel:(float)theWeight { float graphMaxY = 400; //The TOP of the graph float graphMinY = 100; //The BOTTOM of the graph float yOffset = 100; //Graph itself is offset 100 pixels in the Y direction float coordDiff = graphMaxY-graphMinY; //The size in pixels of the graph float weightDiff = self.highestWeight-self.lowestWeight; //The weight gap float pixelIncrement = coordDiff/weightDiff; float weightY = (theWeight*pixelIncrement)-(coordDiff-yOffset); //The return value return weightYpixel; }

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  • Call other activities in an activity?

    - by Mohit Deshpande
    Say I have 2 activities (ActivityOne and ActivityTwo). How would I call ActivityTwo from ActivityOne? Then how would I return to ActivityOne from ActivityTwo? For example, I have a listview with all the contacts on the host phone. When I tap on a contact, another activity shows information and allows editing of that contact. Then I could hit the back button, and I would go back to the exact state that ActivityOne was in before I called ActivityTwo. I was thinking an Intent object, but I am not sure. Could someone post some code?

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  • Spring singleton lifecycle

    - by EugeneP
    Reading this When a bean is a singleton, only one shared instance of the bean will be managed and all requests for beans with an id or ids matching that bean definition will result in that one specific bean instance being returned. Will be managed... What does that mean? If there's only one object, than any modification to this object will result in that every another attempt to get this bean will return a modified instance??

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  • Specifying Locations relative to the context XML in Spring

    - by Daisuke Shimamoto
    Is there a way to reference a properties file relative to the Spring's context file itself? What I want to do is the below: <bean class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer"> <property name="locations"> <list> <!-- I want the below to be relative to this context XML file. --> <value>app.properties</value> </list> </property> </bean> I was imagining something like ${contextpath} that I could prepend to the above "app.properties" but couldn't find anything helpful. Thanks.

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  • Android Convert Central Time to Local Time

    - by chedstone
    I have a MySql database that stores a timestamp for each record I insert. I pull that timestamp into my Android application as a string. My database is located on a server that has a TimeZone of CST. I want to convert that CST timestamp to the Android device's local time. Can someone help with this?

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  • Is map/collection order stable between calls?

    - by John
    If I have a hash map and iterate over the objects repeatedly, is it correct that I'm not guaranteed the same order for every call? For example, could the following print two lines that differ from each other: Map<String,Integer> map = new HashMap<String,Integer>() {{ put("a", 1); put("b", 2); put("c", 3); }}; System.out.println(map); System.out.println(map); And is this the case for sets and collections in general? If so, what's the best way in case you have to iterate twice over the same collection in the same order (regardless of what order that is)? I guess converting to a list.

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