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Java Network Q&A » url » inputstream 

1. Java resource closing    stackoverflow.com

I'm writing an app that connect to a website and read one line from it. I do it like this:

try{
        URLConnection connection = new ...

2. javax.imageio.IIOException: Can't get input stream from URL!    stackoverflow.com

I write a code in java code to save images from specified links. It works fine mostly, but some time it throws an exception ie "javax.imageio.IIOException: Can't get input stream from URL! at ...

3. Trying to read from a URL(in Java) produces gibberish on certain occaisions    stackoverflow.com

I'm trying to read from a URL, and then print the result.

BufferedReader in = new BufferedReader(
     new InputStreamReader(new URL("http://somesite.com/").openStream(), "UTF-8"));
String s = "";
while ((s=in.readLine())!=null) System.out.println(s);
in.close();
It works great ...

4. Getting MalformedURLException when using inputStream in JUnit    stackoverflow.com

So I'm writing some unit tests using JUnit for an RSS parser application. The part I'm trying to test takes a URL and opens a connection to it (HttpURLConnection), then ...

5. Gunzipping Contents of a URL - Java    stackoverflow.com

So as the title suggests, I'm trying to get and gunzip a string from an HTTP request.

urlConn = url.openConnection();
int len = CONTENT_LENGTH
byte[] gbytes = new byte[len];
gbuffer = new GZIPInputStream(urlConn.getInputStream(), len);
System.out.println(gbuffer.read(gbytes)+"/"+len);
System.out.println(gbytes);
result = ...

6. InputStream from a file url    stackoverflow.com

How do I get an InputStream from a file url? for example, I want to take the file at the url wwww.somewebsite.com/a.txt and read it as an InputStream in Java, through a ...

7. Java: HttpComponents gets rubbish Response from input Stream from a specific URL    stackoverflow.com

I am currently trying to get HttpComponents to send HttpRequests and retrieve the Response. On most URLs this works without a problem, but when I try to get the URL of a ...

8. 2 BufferedReaders from one InputStream    stackoverflow.com

I made the following code:

try {
  URL url = new URL("http://bbc.com");
  is = url.openStream();
  BufferedReader in = new BufferedReader(new InputStreamReader(is, "UTF-8"));
  System.out.println(in.readLine());
  //in.close(); with this next ...

9. Best way to log java.net.URL results    stackoverflow.com

I have a library function as follows:

public static InputStream getResource(String url) throws MalformedURLException,
    IOException {
    return new URL(url).openConnection().getInputStream();
}
What would be the best way to add ...

10. How to tell how far along an InputStream a loop is in Java    stackoverflow.com

I'm trying to make a downloader so I can automatically update my program. The following is my code so far:

  public void applyUpdate(final CharSequence ver)
  {
    ...

11. URL stream throws exception when trying to read from input stream    coderanch.com

Hi guys, I tried reading from an input stream which returned a 401. Actually, the server side returns 401 whenever the application server cannot fulfill the client's request, however the 401 contains a message which I will like to read. I have tried this with mozilla and the return message is properly displayed even at 401. How can I read this ...

12. occasional NullPointerException in InputStream (java.net.URL)    coderanch.com

The below code works well, but occasionally, there is a NullPointerException after "logger.debug("after read:"+numRead);"(here can always print the number, for example, if normal response is "abc", it will return 3, but if exception, I can also get 3 in that line). Any master helps to point out whether there is any potential exception in some line with relevant possible reason, thanks ...

13. get image from URL using inputstream    coderanch.com

Hi, I am trying to get images from a url using a inputstream. The following is the sample code. int length=connection.getContentLength(); //System.out.println("Content length : "+length); inputStream=connection.getInputStream(); if (inputStream!=null) { blob=new byte[length]; inputStream.read(blob); inputStream.close(); inputStream=null; } I am able to get the image but only part of the image is good and the other part is damaged(Some thing like a mosaic type). ...

14. inputstream reading wrong file when a url is passed.    coderanch.com

hi all, i am working on a web project. now the task is as follows: i have a webpage on hosted on localhost. i have a page which i upload to it and put it in a specified location using a servlet. now i have a java class which has to read this uploaded file, i get the path of the ...

16. Get more than 1 URL from inputStream    forums.oracle.com

17. How to extract URL from inputStream    forums.oracle.com

You have two options. You can either parse the whole file, preferably using an existing HTML parser. You get back what's known as a "parse tree", representing the HTML document abstractly. Then you look through that data structure for the data you want. This is probably overkill for this task. Or you can read the file, line-by-line (you don't have to ...

18. InputStream through URL ,problem while rendering the file    forums.oracle.com

ByteArrayInputStream ba = new ByteArrayInputStream(attachment); FileOutputStream fos=null; try { fos = new FileOutputStream(file); } catch (FileNotFoundException e1) { // TODO Auto-generated catch block e1.printStackTrace(); } int n=0; while((n = ba.read()) >= 0) { try { fos.write(attachment,0,n); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } } try { fos.close(); } catch (IOException e1) { // TODO Auto-generated catch ...

19. Reading from URL InputStream    forums.oracle.com

After creation byte array result[] it has size 18811. But readResult variable has value which less then result.length. Why? Because the implementation of the read method doesn't say it will guarantee to read the whole enchilada at once. There's nothing stopping you from reading a chunk at a time in a loop though. And you shouldn't create an arbitrarily-large array. What ...

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