java.net.URL read stream to byte[]

Just extending Barnards's answer with commons-io. Separate answer because I can not format code in comments.

InputStream is = null;
try {
  is = url.openStream ();
  byte[] imageBytes = IOUtils.toByteArray(is);
}
catch (IOException e) {
  System.err.printf ("Failed while reading bytes from %s: %s", url.toExternalForm(), e.getMessage());
  e.printStackTrace ();
  // Perform any other exception handling that's appropriate.
}
finally {
  if (is != null) { is.close(); }
}

http://commons.apache.org/io/api-1.4/org/apache/commons/io/IOUtils.html#toByteArray(java.io.InputStream)

There's no guarantee that the content length you're provided is actually correct. Try something akin to the following:

ByteArrayOutputStream baos = new ByteArrayOutputStream();
InputStream is = null;
try {
  is = url.openStream ();
  byte[] byteChunk = new byte[4096]; // Or whatever size you want to read in at a time.
  int n;

  while ( (n = is.read(byteChunk)) > 0 ) {
    baos.write(byteChunk, 0, n);
  }
}
catch (IOException e) {
  System.err.printf ("Failed while reading bytes from %s: %s", url.toExternalForm(), e.getMessage());
  e.printStackTrace ();
  // Perform any other exception handling that's appropriate.
}
finally {
  if (is != null) { is.close(); }
}

You'll then have the image data in baos, from which you can get a byte array by calling baos.toByteArray().

This code is untested (I just wrote it in the answer box), but it's a reasonably close approximation to what I think you're after.

Here's a clean solution:

private byte[] downloadUrl(URL toDownload) {
    ByteArrayOutputStream outputStream = new ByteArrayOutputStream();

    try {
        byte[] chunk = new byte[4096];
        int bytesRead;
        InputStream stream = toDownload.openStream();

        while ((bytesRead = stream.read(chunk)) > 0) {
            outputStream.write(chunk, 0, bytesRead);
        }

    } catch (IOException e) {
        e.printStackTrace();
        return null;
    }

    return outputStream.toByteArray();
}