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• Reading http file on url

  - by Nkunzis

  I am trying to read an xml file on an http url. "Unexpected end of file from server" keeps on coming . the page is password protected, I would like to know if I am properly giving in my url authentication details. I tried with non protected pages and I can read them properly. Here is my code: URL url = new URL("http://username:[email protected]:0000/test.xml"); URLConnection yc = url.openConnection(); BufferedReader in = new BufferedReader(new InputStreamReader(yc.getInputStream())); String inputLine; while ((inputLine = in.readLine()) != null) System.out.println(inputLine); in.close();

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• how to get values sent by location.replace(URL)

  - by kawtousse

  hi everyone, In my javascript function I do livke this in order to redirect parameters to servlet: var ids1=document.getElementById("projet").value; document.location.href("http://localhost:8080/Opc_Web_App/ServletAffectation?ids1="+ids1); and in the servlet I do the following to get Value: String idprojet= request.getParameter("projet"); System.out.println("le projet selectionné est :" +idprojet); the problem that i didnt have the result of System.out.print in my screen; so in other terms the servlet didn't get the parameter. I can not see the problemn untill now. Please help. Thank you.

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• Send value to servlet with javascript

  - by kawtousse

  hi everyone, In my JavaScript function I do like this in order to redirect parameters to servlet: var ids1=document.getElementById("projet").value; document.location.href("http://localhost:8080/Opc_Web_App/ServletAffectation?ids1="+ids1); and in the servlet I do the following to get Value: String idprojet= request.getParameter("ids1"); System.out.println("le projet selectionné est :" +idprojet); the problem that i didn't have the result of System.out.print in my screen; so in other terms the servlet didn't get the parameter. I can not see the problem until now. Please help. Thank you.

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• Can a thread call wait() on two locks at once in Java (6)

  - by Dr. Monkey

  I've just been messing around with threads in Java to get my head around them (it seems like the best way to do so) and now understand what's going on with synchronize, wait() and notify(). I'm curious about whether there's a way to wait() on two resources at once. I think the following won't quite do what I'm thinking of: synchronized(token1) { synchronized(token2) { token1.wait(); token2.wait(); //won't run until token1 is returned System.out.println("I got both tokens back"); } } In this (very contrived) case token2 will be held until token1 is returned, then token1 will be held until token2 is returned. The goal is to release both token1 and token2, then resume when both are available (note that moving the token1.wait() outside the inner synchronized loop is not what I'm getting at). A loop checking whether both are available might be more appropriate to achieve this behaviour (would this be getting near the idea of double-check locking?), but would use up extra resources - I'm not after a definitive solution since this is just to satisfy my curiosity.

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• scala coalesces multiple function call parameters into a Tuple -- can this be disabled?

  - by landon9720

  This is a troublesome violation of type safety in my project, so I'm looking for a way to disable it. It seems that if a function takes an AnyRef (or a java.lang.Object), you can call the function with any combination of parameters, and Scala will coalesce the parameters into a Tuple object and invoke the function. In my case the function isn't expecting a Tuple, and fails at runtime. I would expect this situation to be caught at compile time. object WhyTuple { def main(args: Array[String]): Unit = { fooIt("foo", "bar") } def fooIt(o: AnyRef) { println(o.toString) } } Output: (foo,bar)

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• C++ string how to

  - by typoknig

  This is a very simple question and I feel stupid for asking it, but I am pressed for time and I need to figure it out :) I just need to know how to make a string that contains text and other variables. For instance in Java I can just do this: String someString; for(int i = 0; i>10; i++){ someString = ("this text has printed " + i + " times"); //how do I create this line in C++? System.out.println(someString); i++; }

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• How to read output of android process command

  - by kevdliu

  I am trying to get the output of android shell command 'getprop' with java since getprop() always returns null no matter what. I tried this from developer.android.com: Process process = null; try { process = new ProcessBuilder() .command("/system/bin/getprop", "build.version") .redirectErrorStream(true) .start(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } InputStream in = process.getInputStream(); //String prop = in.toString(); System.out.println(in); process.destroy(); However what is printed is not the output but a bunch of characters and numbers (dont have the exact output right now). How can i get the output of the process? Thanks!

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• inputMismatchException Java reading doubles from plain text file

  - by user939287

  Using double variable = inputFile.nextDouble(); Gives the mismatch error and I can't figure out why... Anyone know what's up? The input file is just a bunch of doubles like 5.0... Okay here is the code snippet String fileName; Scanner scanner = new Scanner(System.in); System.out.println("\nEnter file name that contains the matrix and vector: "); fileName = scanner.nextLine(); Scanner inputFile = new Scanner(fileName); double a1 = inputFile.nextDouble(); the input file is a plain text document .txt in this format 5.0 4.0 -3.0 4.0 2.0 5.0 6.0 5.0 -2.0 -13.0 4.0 12.0 I don't understand why it wouldn't take those as doubles... As far as what its expecting the format of the file to be... I suppose binary? isn't that the default? I didn't specify in the code...

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• Get _id from MongoDB using Scala

  - by user2438383

  For a given JSON how do I get the _id to use it as an id for inserting in another JSON? Tried to get the ID as shown below but does not return correct results. private def getModelRunId(): List[String] = { val resultsCursor: List[DBObject] = modelRunResultsCollection.find(MongoDBObject.empty, MongoDBObject(FIELD_ID -> 1)).toList println("resultsCursor >>>>>>>>>>>>>>>>>> " + resultsCursor) resultsCursor.map(x => (Json.parse(x.toString()) \ FIELD_ID).asOpt[String]).flatten } { "_id": ObjectId("5269723bd516ec3a69f3639e"), "modelRunId": ObjectId("5269723ad516ec3a69f3639d"), "results": [ { "ClaimId": "526971f5b5b8b9148404623a", "pricingResult": { "TxId": 0, "ClaimId": "Large_Batch_1", "Errors": [ ], "Disposition": [ { "GroupId": 1, "PriceAmt": 20, "Status": "Priced Successfully", "ReasonCode": 0, "Reason": "RmbModel(PAM_DC_1):ProgramNode(Validation CPG):ServiceGroupNode(Medical Services):RmbTerm(RT)", "PricingMethodologyId": 2, "Lines": [ { "Id": 1 } ] } ] } },

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• Why Timer does not work if we do not generate a window?

  - by Roman

  Here is the code: import java.awt.event.ActionEvent; import java.awt.event.ActionListener; import javax.swing.JFrame; import javax.swing.Timer; public class TimerSample { public static void main(String args[]) { new JFrame().setVisible(true); ActionListener actionListener = new ActionListener() { public void actionPerformed(ActionEvent actionEvent) { System.out.println("Hello World Timer"); } }; Timer timer = new Timer(500, actionListener); timer.start(); } } It generates a window and then periodically prints "Hello World Timer" in the terminal (Command Prompt). If I comment this line new JFrame().setVisible(true); the application do not print anything to the command line. Why?

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• the PrintWriter servlet Buffer and displayind data from a jsp

  - by nabilaloui

  Hello all, I need really your help please. What I do is to build a table in html tags in my servlet then when trying to send this table to a servlet for the display using: Response.sendRedirect this did not work. I have an error but I don't know the cause. I search to how do it since I use: response.setContentType("text/html"); PrintWriter out = response.getWriter(); out.println("<...");.... thinks a lot for your help

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• Send a String[] ArrayList over Socket connection

  - by Duncan Palmer

  So i'm trying to send a String[] Array/List over an open socket connection. I currently have this code: Sending: public void sendData() { try { OutputStream socketStream = socket.getOutputStream(); ObjectOutputStream objectOutput = new ObjectOutputStream(socketStream); objectOutput.writeObject(new String[] {"Test", "Test2", "Test3"}); objectOutput.close(); socketStream.close(); } catch (Exception e) { System.out.println(e.toString()); } } Recieving: public Object readData() { try { InputStream socketStream = socket.getInputStream(); ObjectInputStream objectInput = new ObjectInputStream(new GZIPInputStream(socketStream)); Object a = objectInput.readObject(); return a; } catch(Exception e) { return null; } } After I have recieved the String array/list on the other end I want to be able to iterate through it like I would do normally so I can get the values. My current code doesn't seem to works as it returns null as the value. is this possible?

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• How do you send an array as part of an (jquery) ajax request

  - by Ankur

  I tried to send an array as part of an ajax request like this: var query = []; // in between I add some values to 'query' $.ajax({ url: "MyServlet", data: query, dataType: "json", success: function(noOfResults) { alert(noOfResults); } }); } I wanted to see what I get back in the servlet, so I used this line: System.out.println(request.getParameterMap().toString()); Which returned {} suggesting an empty map. Firebug tells me I am getting a 400 bad request error If I send a queryString like attribute=value as the 'data' then everything works fine, so it has to do with not being able to send an array as is. What do I have to do to get that data into the servlet for further processing. I don't want to pull it out and turn it into a queryString in the JS if I can avoid it. EDIT: I used the .serializeArray() (jQuery) function before sending the data. I don't get the 400 but nothing useful is being sent through.

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• removing the first value in an array c# or java

  - by MrCode

  hey there i was working on a program and was thinking is it possible was to remove the value from the first element in an array. Has anyone any ideas on how this could be done ? thanks for all input is much appreciated. i have only tried removing from the last element wasnt sure on how i would remove the first this is how i done the last element try { if (isEmpty()) { throw new Exception("list is empty"); } size = size -1; return values[size]; } catch(Exception e) { System.out.println(e); return -1; }

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• List files starting with a specific name using java

  - by user3610075

  i want to list files starting with a name like "Report" from a folder. i found this in google to list all files but i don't how to list file starting with a name. Thank you File directory = new File("C:\\Users\\kiki\\Downloads"); File[] files = directory.listFiles(); for (int index = 0; index < files.length; index++) { //Print out the name of files in the directory System.out.println(files[index].toString()); }

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• Set intersection of two strings

  - by user1785712

  import java.util.*; class set { public static void main(String args[]) { TreeSet<Character> t1 = new TreeSet<Character>(); TreeSet<Character> t2 = new TreeSet<Character>(); String s1 = "Ambitab bachan"; String s2 = "Ranjikanth"; for(char c1:s1.toCharArray()) t1.add(c1); for(char c2:s2.toCharArray()) t2.add(c2); t2.retainAll(t1); System.out.println(t2); } } this program find the common character in two different string. in this program Treeset is used to store the value and retainAll() method is used to find the common characters. can anybody help me reduce the line of coding.thanks in advance

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• Cant communicate with server in java

  - by cerq

  i m trying to write server to client program but i cant communicate with server in java. Below there is code block in my main. InetAddress addr = InetAddress.getLocalHost(); ipAddress = "78.162.206.164"; ServerSocket serverSocket = new ServerSocket(0); String randomStringForPlayerName = RandomStringGenerator.generateRandomString(); baseForReqOpp += ipAddress + " " + serverSocket + " " + randomStringForPlayerName; Socket socket = new Socket(host,2050); socket.setSoTimeout(100); in = new BufferedReader(new InputStreamReader(socket.getInputStream())); out = new PrintWriter(socket.getOutputStream()); out.write(baseForReqOpp); out.flush(); System.out.println(in.read()); i know there is no problem in server code and all the ports in communication are ok. But i cant read anything from server. What can be the problem?

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• Passing a generic class to a java function?

  - by rodo

  Is it possible to use Generics when passing a class to a java function? I was hoping to do something like this: public static class DoStuff { public <T extends Class<List>> void doStuffToList(T className) { System.out.println(className); } public void test() { doStuffToList(List.class); // compiles doStuffToList(ArrayList.class); // compiler error (undesired behaviour) doStuffToList(Integer.class); // compiler error (desired behaviour) } } Ideally the List.class and ArrayList.class lines would work fine, but the Integer.class line would cause a compile error. I could use Class as my type instead of T extends Class<List> but then I won't catch the Integer.class case above.

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• question about random select

  - by davit-datuashvili

  here is code print number in decreasing order import java.util.*; public class select { public static void Select(int m,int n){ Random r=new Random(); if (m>0) if (r.nextInt(0x3fff8001) % n <m ){ System.out.println(n-1); Select(m-1,n-1); } else{ Select(m,n-1); } } public static void main(String[]args){ int m=35; int n=200; Select(m,n); } } and question is how to changes code such that print number in increasing order? please help

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• What does this java output mean?!

  - by Phil

  public class Arrys { private int[] nums; //Step 3 public Arrys (int arrySize) { nums = new int[arrySize]; } public int [] getNums (){ return nums; } } Test class: public class TestArrys { public static void main(String args[]) { //Step 4 Arrys arry = new Arrys(10); System.out.println("\nStep4 "); for(int index = 0; index < arry.getNums().length; index++) { System.out.print(arry.getNums()); } } } It's incredibly simple, that is why I think I'm doing something fundamentally wrong. All I want is to display the value of the array. This is what I get back. I am totally lost, there is nothing in my book that explains this nor does googling it help. Step4 [I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440

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• How to validate phone number(US format) in Java?

  - by Maxood

  I just want to know where am i wrong here: import java.io.*; class Tokens{ public static void main(String[] args) { //String[] result = "this is a test".split(""); String[] result = "4543 6546 6556".split(""); boolean flag= true; String num[] = {"0","1","2","3","4","5","6","7","8","9"}; String specialChars[] = {"-","@","#","*"," "}; for (int x=1; x<result.length; x++) { for (int y=0; y<num.length; y++) { if ((result[x].equals(num[y]))) { flag = false; continue; } else { flag = true; } if (flag == true) break; } if (flag == false) break; } System.out.println(flag); } }

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• How do I setup Eclipse to stop on the line an exception occured?

  - by Chris Persichetti

  Hi, How can I setup Eclipse to stop at the point an exception occurred. I have an Eclipse breakpoint setup to break on an exception. In the code example below, the problem I'm having is Eclipse tries to open the Integer source code. Is there any way to just have debugger break at the point shown in my code example? If I move down the stack trace, I will get to this line, it'd be nice if there's a way to do this without the "Source not found" window coming up. This can be done in Visual Studio, so it's driving me crazy not being able to find a way to do this in Eclipse. package com.test; public class QuickTest { public static void main(String[] args) { try { test(); } catch(NumberFormatException e) { System.out.println(e.getMessage()); } } private static void test() { String str = "notAnumber"; Integer.parseInt(str);//<----I want debugger to stop here } }

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• Why javabeans framework create the IndexedPropertyDescriptor for the NON index method

  - by George Macus

  I'm not familiar with java beans framework, in the below scenario, I got the IndexedPropertyDescriptor for the method getFooWithX, could someone explain why? public class IntrospectorTest { public static void main(String[] args) throws IntrospectionException { BeanInfo info = Introspector.getBeanInfo(SubClass.class); PropertyDescriptor[] descriptors = info.getPropertyDescriptors(); for (int i = 0; i < descriptors.length; i++) { System.out.println(descriptors[i].getClass().getName() + ":" + descriptors[i].getName()); } } } abstract class BaseClass { public abstract Object getFoo(); } abstract class SubClass extends BaseClass { public Object getFooWithX(int x) { return null; } } and the result will be: java.beans.PropertyDescriptor:class java.beans.PropertyDescriptor:foo java.beans.IndexedPropertyDescriptor:fooWithX Why?

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• getHeaderField("WWW-Authenticate") giving improper value in J2ME

  - by kehers

  I'm building a client for an api that uses http digest access authentication for authentication. I have studied the rfc to know the setup the required response headers and this works well on my emulator. Problem however is when I test on my phone (nokia E5), I found out that getting the www-authenticate header from the returned headers doesnt get the full value [code] // c = (HttpConnection) Connector.open(url) and other declarations String digest = c.getHeaderField("WWW-Authenticate"); System.out.println(digest); // gives only: Digest //no realm, qop and others [/code] I'm I doing something wrong or it is from the phone? What are my other options?

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• How to check null element if it is integer array in Java?

  - by masato-san

  I'm quite new to Java and having an issue checking null element in integer array. I'm using Eclipse for editor and the line that checks null element is showing error: Line that complains: if(a[i] != null) { Error msg from Eclipse: The operator != is undefined for the argument type(s) int, null In PHP, this works without any problem but in Java it seems like I have to change the array type from integer to Object to make the line not complain (like below) Object[] a = new Object[3]; So my question is if I still want to declare as integer array and still want to check null, what is the syntax for it? Code: public void test() { int[] a = new int[3]; for(int i=0; i<a.length; i++) { if(a[i] != null) { //this line complains... System.out.println('null!'); } } }

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