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 >

  • problem in jdbc preparestatement

    - by akshay
    i am geting error when i try to use following,why is it so? ResultSet findByUsername(String tablename,String field,String value) { pStmt = cn.prepareStatement("SELECT * FROM" + tablename +" WHERE ? = ? "); pStmt.setString(1, tablename); pStmt.setString(2,field); pStmt.setString(3,value); return(pStmt.executeQuery()); } also i tried following , but its not working too ResultSet findByUsername(String tablename,String field,String value) { String sqlQueryString = " SELECT * FROM " + tablename +" WHERE " + filed + "= ? ") cn.prepareStatement(sqlQuery); pStmt.setString(1, value); return(pStmt.executeQuery()); }

    Read the article

  • How can we compute the LAST page with JPA?

    - by Kevin
    I would like to implement pagination in my Servlet/EJB/JPA-Hibernate project, but I can't figure out how only one page from the query and know the number of pages I must display I use setFirstResult(int first) ; setMaxResults(int max) ; and that's working alright, but how can I know how many pages I will have in total? (Hibernate is my JPA provider, but I would prefer using only JPA if possible) UPDATE: COUNT() seems to be the better/easiest solution; but what can be the cost of SELECT COUNT(*) FROM ... in comparison with executeQuery("SELECT * FROM ...).getListResult().size() ?

    Read the article

  • Is it possible to Hide Text through a Style?

    - by Sandro
    I currently have a JTextPane that will be displaying text coming in from different streams. The way that the user can tell which stream the text came from is that the text from each stream has a different Style to it. Is there a way to make a Style that will hide the text so that I can filter out different pieces of text? Thank you.

    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

  • Is it possible to draw simultaneously on a panel?

    - by swift
    I have to develop a whiteboard application in which both the local user and the remote user should be able to draw simultaneously, is this possible? If possible then any logic? I have already developed a code but in which i am not able to do this, when the remote user starts drawing the shape which i am drawing is being replaced by his shape and co-ordinates. This problem is only when both draw simultaneously. any idea guys?

    Read the article

  • How to persist in order

    - by parhs
    I have two entities: EXAM and EXAM_NORMAL. EXAM @Id @GeneratedValue(strategy=GenerationType.IDENTITY) private Long id; private String name; private String codeName; @Enumerated(EnumType.STRING) private ExamType examType; @ManyToOne private Category category; @OneToMany(mappedBy="id",cascade=CascadeType.PERSIST) @OrderBy("id") private List<Exam_Normal> exam_Normal; EXAM_NORMAL @Id private Long item; @Id @ManyToOne private Exam id; @Enumerated(EnumType.STRING) private Gender gender; private Integer age_month_from; private Integer age_month_to; The problem is that if I put a list of EXAM_NORMAL at an EXAM class if I try to persist(EXAM) I get an error because it tries to persist EXAM_NORMAL first but it cant because the primary key of EXAM is missing because it isn't persisted... Is there any way to define the order? Or should I set null the list, persist and then set the list again? thanks:)

    Read the article

  • Reuse Hibernate session in thread

    - by Marco
    Hello, I've read somewhere that when a session is flushed or a transaction is committed, the session itself is closed by Hibernate. So, how can i reuse an Hibernate Session, in the same thread, that has been previously closed? Thanks

    Read the article

  • Allow Incorrect Package Name in Eclipse

    - by SamBeran
    I have some classes in my current project which have the wrong package declaration (they are in the wrong folder for their declared package.) Unfortunately, fixing the problem by moving the class is not an option. Is there a way I can get eclipse to ignore the error?

    Read the article

  • Do running times match with O(nlogn)?

    - by user472221
    Hi I have written a class(greedy strategy) that at first i used sort method which has O(nlogn) Collections.sort(array, new SortingObjectsWithProbabilityField()); and then i used the insert method of binary search tree which takes O(h) and h here is the tree height. for different n ,the running time will be : n,running time 17,515428 33,783340 65,540572 129,1285080 257,2052216 513,4299709 which I think is not correct because for increasing n , the running time should almost increase. This method will take the running time: Exponent = -1; for(int n = 2;n<1000;n+=Math.pow(2,exponent){ for (int j = 1; j <= 3; j++) { Random rand = new Random(); for (int i = 0; i < n; i++) { Element e = new Element(rand.nextInt(100) + 1, rand.nextInt(100) + 1, 0); for (int k = 0; k < i; k++) { if (e.getDigit() == randList.get(k).getDigit()) { e.setDigit(e.getDigit() + 1); } } randList.add(e); } double sum = 0.0; for (int i = 0; i < randList.size(); i++) { sum += randList.get(i).getProbability(); } for (Element i : randList) { i.setProbability(i.getProbability() / sum); } //Get time. long t2 = System.nanoTime(); GreedyVersion greedy = new GreedyVersion((ArrayList<Element>) randList); long t3 = System.nanoTime(); timeForGreedy = timeForGreedy + t3 - t2; } System.out.println(n + "," + "," + timeForGreedy/3 ); exponent++; } thanks

    Read the article

  • how can this inner enum code be improved ?

    - by mafalda
    I have this construct public class Constants{ enum SystemA implements Marker{ ConstOne(1), ConstTwo(2), ConstThree(3); SystemA(int i) { number =i; } int number; } enum SystemB implements Marker{ ConstFour(4), ConstFive(5), ConstSix(6); SystemB(int i) { number =i; } int number; } } I have Marker so I can pass to method like this method(Constants.SystemA) or method(Constants.SystemB) What is the best way to list all the enum values ? I also want to make sure that it is not duplicating the number in any of the enums

    Read the article

  • How to sort an array by (smallest, largest, second smallest, second, largest) etc?

    - by Binka
    Any ideas? I can sort an array. But not in this pattern? It needs to sort by the pattern I mentioned above. public void wackySort2(int[] nums) { int sign = 0; int temp = 0; int temp2 = 0; for (int i = 0; i < nums.length; i++) { for (int j = 0; j < nums.length - 1; j++) { if (nums[j] > nums[j + 1]) { temp = nums[j]; nums[j] = nums[j + 1]; nums[j + 1] = temp; //sign = 1; System.out.println("Something has been done"); } } } }

    Read the article

  • Possible to skip track from an Android application?

    - by parse
    I'm planning on doing a application for Android 2.1 that changes song every minute (through what I hope exists in Android, "next") for the application using the audio device atm. So if I have Spotify running in background already, playing music, can I through my program change to the next track? Let me know if I was unclear about anything. Thanks in advance!

    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

  • Rotated image in ImageView

    - by Lars D
    I want to show an arrow that indicates the direction towards a goal, using the orientation sensor and current GPS position. Everything works well, except that I want to rotate the arrow image in my ImageView. The current code, which shows the arrow pointing upwards, is this: ImageViewArrow.setImageResource(R.drawable.arrow); What is the best solution for showing the arrow, rotated by N degrees?

    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

  • GWT + XML documents with namespaces

    - by chris_l
    I'd like to get a quick overview of available solutions (libraries, ...) that allow me to work with XML documents with namespaces on a DOM level - in GWT's client side. Additionally, I'm looking for an XPath solution that can work on that DOM (even if it requires writing my own XPath Navigator). XML parsing and serialization isn't necessary on the client - this can be done on the server.

    Read the article

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