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HTTP Post multipart file upload in Java ME

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Created: jappit (12 May 2008)
Last edited: hamishwillee (25 Jul 2013)

Here is a J2ME class to handle file uploads via HTTP POST Multipart Requests.

Source Code: HttpMultipartRequest class

import java.io.ByteArrayOutputStream;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.Enumeration;
import java.util.Hashtable;
 
import javax.microedition.io.Connector;
import javax.microedition.io.HttpConnection;
 
public class HttpMultipartRequest
{
static final String BOUNDARY = "----------V2ymHFg03ehbqgZCaKO6jy";
 
byte[] postBytes = null;
String url = null;
 
public HttpMultipartRequest(String url, Hashtable params, String fileField, String fileName, String fileType, byte[] fileBytes) throws Exception
{
this.url = url;
 
String boundary = getBoundaryString();
 
String boundaryMessage = getBoundaryMessage(boundary, params, fileField, fileName, fileType);
 
String endBoundary = "\r\n--" + boundary + "--\r\n";
 
ByteArrayOutputStream bos = new ByteArrayOutputStream();
 
bos.write(boundaryMessage.getBytes());
 
bos.write(fileBytes);
 
bos.write(endBoundary.getBytes());
 
this.postBytes = bos.toByteArray();
 
bos.close();
}
 
String getBoundaryString()
{
return BOUNDARY;
}
 
String getBoundaryMessage(String boundary, Hashtable params, String fileField, String fileName, String fileType)
{
StringBuffer res = new StringBuffer("--").append(boundary).append("\r\n");
 
Enumeration keys = params.keys();
 
while(keys.hasMoreElements())
{
String key = (String)keys.nextElement();
String value = (String)params.get(key);
 
res.append("Content-Disposition: form-data; name=\"").append(key).append("\"\r\n")
.append("\r\n").append(value).append("\r\n")
.append("--").append(boundary).append("\r\n");
}
res.append("Content-Disposition: form-data; name=\"").append(fileField).append("\"; filename=\"").append(fileName).append("\"\r\n")
.append("Content-Type: ").append(fileType).append("\r\n\r\n");
 
return res.toString();
}
 
public byte[] send() throws Exception
{
HttpConnection hc = null;
 
InputStream is = null;
 
ByteArrayOutputStream bos = new ByteArrayOutputStream();
 
byte[] res = null;
 
try
{
hc = (HttpConnection) Connector.open(url);
 
hc.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + getBoundaryString());
 
hc.setRequestMethod(HttpConnection.POST);
 
OutputStream dout = hc.openOutputStream();
 
dout.write(postBytes);
 
dout.close();
 
int ch;
 
is = hc.openInputStream();
 
while ((ch = is.read()) != -1)
{
bos.write(ch);
}
res = bos.toByteArray();
}
catch(Exception e)
{
e.printStackTrace();
}
finally
{
try
{
if(bos != null)
bos.close();
 
if(is != null)
is.close();
 
if(hc != null)
hc.close();
}
catch(Exception e2)
{
e2.printStackTrace();
}
}
return res;
}
}

Sample usage

Here's a code snippet to upload a file via HttpMultipartRequest class:

byte[] fileBytes = getFileBytes(); //retrieve file bytes with your own code
 
Hashtable params = new Hashtable();
params.put("custom_param", "param_value");
params.put("custom_param2", "param_value2");
 
HttpMultipartRequest req = new HttpMultipartRequest(
"http://www.server.com/uploadScript.php",
params,
"upload_field", "original_filename.png", "image/png", fileBytes
);
 
byte[] response = req.send();

Sample server code (PHP)

This is a sample PHP script that handles the upload. It doesn't actually save the uploaded file, but only displays some infos about the upload size and parameters.

<?php
 
$filesize = filesize($_FILES['upload_field']['tmp_name']);
 
echo "The uploaded file size is " . $filesize . " bytes\n";
 
foreach($_POST as $key => $value)
{
echo "Parameter name: " . $key . ", value: " . $value . "\n";
}
 
?>
This page was last modified on 25 July 2013, at 09:57.
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