Reputation: 33
I am trying to download an image from a server through sockets. My code works fine, but when I download the image, the size is correct but the image does not open. I don't know what am I doing wrong. Any suggestion? Thank you
Socket socket = new Socket(servername, 80);
DataOutputStream bw = new DataOutputStream(new DataOutputStream(socket.getOutputStream()));
bw.writeBytes("GET "+filename+" HTTP/1.1\n");
bw.writeBytes("Host: "+servername+":80\n\n");
DataInputStream in = new DataInputStream(socket.getInputStream());
OutputStream dos = new FileOutputStream("testtttt.jpg");
int count;
byte[] buffer = new byte[2048];
while ((count = in.read(buffer)) != -1)
{
dos.write(buffer, 0, count);
dos.flush();
}
dos.close();
System.out.println("image transfer done");
socket.close();
}
Upvotes: 3
Views: 5457
Reputation: 123
My reputation is not enough to comment, so have to start a new answer.
ug__'s answer is great, but the line
buffer = string.substring(indexOfEOH+4).getBytes();
has some problems, the buffer will be corrupted. For example,
byte[] before = new byte[]{(byte)0xf1, (byte)0xf2, (byte)0xf3, (byte)0xf4};
String str = new String(before, 0, before.length);
byte[] after = str.getBytes();
before
and after
will not be the same.
So I modified ug__'s code a little bit:
OutputStream dos = new FileOutputStream("test.jpg");
int count, offset;
byte[] buffer = new byte[2048];
boolean eohFound = false;
while ((count = in.read(buffer)) != -1)
{
offset = 0;
if(!eohFound){
String string = new String(buffer, 0, count);
int indexOfEOH = string.indexOf("\r\n\r\n");
if(indexOfEOH != -1) {
count = count-indexOfEOH-4;
offset = indexOfEOH+4;
eohFound = true;
} else {
count = 0;
}
}
dos.write(buffer, offset, count);
dos.flush();
}
in.close();
dos.close();
Upvotes: 1
Reputation: 11440
You need to add a \r before your \n on all requests, additionally you should flush your output stream to the socket.
Socket socket = new Socket(servername, 80);
DataOutputStream bw = new DataOutputStream(socket.getOutputStream());
bw.writeBytes("GET "+filename+" HTTP/1.1\r\n");
bw.writeBytes("Host: "+servername+":80\r\n\r\n");
bw.flush();
Additionally you will get some HTTP response headers with your request. Obiously this is information you dont want in your image, your response will look something like this:
HTTP/1.1 200 OK
Date: Thu, 14 Nov 2013 18:39:47 GMT
Server: Apache/2.4.3 (Win32) OpenSSL/1.0.1c PHP/5.4.7
Accept-Ranges: bytes
ETag: W/"2956-1374616977919"
Last-Modified: Tue, 23 Jul 2013 22:02:57 GMT
Content-Type: image/png;charset=UTF-8
Content-Length: 2956
‰JPG....heres your image data
I just wrote up this method to get rid of the HTTP headers sent across. The idea is to not write any data before a \r\n\r\n occurrence. That sequence represents the end of the header response and any data before that is not our image. I know there is a cleaner way to do it but this way was fast for me to write :)
OutputStream dos = new FileOutputStream("c:\\testtttt.jpg");
int count;
byte[] buffer = new byte[2048];
boolean eohFound = false;
while ((count = in.read(buffer)) != -1)
{
if(!eohFound){
String string = new String(buffer, 0, count);
int indexOfEOH = string.indexOf("\r\n\r\n");
if(indexOfEOH != -1) {
count = count-indexOfEOH-4;
buffer = string.substring(indexOfEOH+4).getBytes();
eohFound = true;
} else {
count = 0;
}
}
dos.write(buffer, 0, count);
dos.flush();
}
in.close();
dos.close();
You can also find another question like yours here: Send HTTP Request manually via socket
Upvotes: 2