Server side code to handle HttpURLConnection in Android - java

I am using HttpURLConnection class to upload text file to my Apache server. For server side code I used php to handle the POST request. I was wondering , how the same thing can be done other then PHP, or the PHP is the best way to handle file uploading on server side?..Also I cannot find any jsp or java code on server side to handle http post request..any code snippet that does same thing other then PHP would be really helpful..Cause I think java or jsp would be easier to code ..Thanks
Android code to upload file:
public void upLoad()
{
String exsistingFileName = path+"//"+"test.txt";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
try {
// ------------------ CLIENT REQUEST
Log.e(Tag, "Inside second Method");
FileInputStream fileInputStream = new FileInputStream(new File(
exsistingFileName));
// open a URL connection to the Servlet
URL url = new URL(urlString);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
// Allow Inputs
conn.setDoInput(true);
// Allow Outputs
conn.setDoOutput(true);
// Don't use a cached copy.
conn.setUseCaches(false);
// Use a post method.
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("Content-Type",
"multipart/form-data;boundary=" + boundary);
DataOutputStream dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: post-data; name=uploadedfile;filename="
+ exsistingFileName + "" + lineEnd);
dos.writeBytes(lineEnd);
Log.e(Tag, "Headers are written");
// create a buffer of maximum size
int bytesAvailable = fileInputStream.available();
int maxBufferSize = 1000;
// int bufferSize = Math.min(bytesAvailable, maxBufferSize);
byte[] buffer = new byte[bytesAvailable];
// read file and write it into form...
int bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);
while (bytesRead > 0) {
dos.write(buffer, 0, bytesAvailable);
bytesAvailable = fileInputStream.available();
bytesAvailable = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// close streams
Log.e(Tag, "File is written");
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
Log.e(Tag, "error: " + ex.getMessage(), ex);
}
catch (IOException ioe) {
Log.e(Tag, "error: " + ioe.getMessage(), ioe);
}
try {
BufferedReader rd = new BufferedReader(new InputStreamReader(conn
.getInputStream()));
String line;
while ((line = rd.readLine()) != null) {
Log.e("Dialoge Box", "Message: " + line);
}
rd.close();
} catch (IOException ioex) {
Log.e("MediaPlayer", "error: " + ioex.getMessage(), ioex);
}
}
Php code on server side :
<?php
$target_path = "./upload/";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>

I think there is no correct answer for this question, PHP can be good as JSP or even ASP. Depends on your platform preference or if you are planning on having a tomcat vs apache server or even a microsoft iis server. Personally I think PHP/Apache is a good choice.

Related

Android upload file to server using php

Hello am trying to upload a file from external storage of android phone to the url in the variable upLoadServerUri as below.
final String upLoadServerUri = "http://www.www.com/UploadToServer.php";
URL url = new URL(upLoadServerUri);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("uploaded_file", fileName);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\""
+ fileName + "\"" + lineEnd);
dos.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
Log.d("Inside ByteRead", "Reading");
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
String serverResponseMessage = conn.getResponseMessage();
fileInputStream.close();
dos.flush();
dos.close();
The code below is being used in the UploadToServer.php file which is the destination url file in the java code above. I don't have knowledge on php end so can you advise me if below code would be able to store data when we trigger a file to it based on the java code above. Thanks.
<?php
// Where the file is going to be placed
$target_path = "recorded/";
/* Add the original filename to our target path.
Result is "uploads/filename.extension" */
$target_path = $target_path . basename( $_FILES['uploaded_file']['name']);
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploaded_file']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
echo "filename: " . basename( $_FILES['uploaded_file']['name']);
echo "target_path: " .$target_path;
}
?>
Please try retrofit. Its simple and easy. Read this tutorial. Upload Files to Server

Posting File to WebServer with Java

I'm trying to post a file to a ASP.NET WEB API (C#) Server with a local Java application.
Basically I'm trying to reproduce the following HTML code in Java SE:
<form name="form1" method="post" enctype="multipart/form-data" action="http://localhost:50447/api/files/">
<div>
<label for="image1">Image File</label>
<input name="image1" type="file" />
</div>
<div>
<input type="submit" value="Submit" />
</div>
What's the simplest way to do this? I'd like to avoid using Apache..
Something like:
String urlToConnect = "http://localhost:50447/api/files/";
String paramToSend = "";
File fileToUpload = new File("C:/Users/aa/Desktop/sample_signed.pdf");
String boundary = Long.toHexString(System.currentTimeMillis()); // Just generate some unique random value.
URLConnection connection = null;
try {
connection = new URL(urlToConnect).openConnection();
} catch (IOException e2) {
// TODO Auto-generated catch block
e2.printStackTrace();
}
connection.setDoOutput(true); // This sets request method to POST.
connection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
PrintWriter writer = null;
try {
writer = new PrintWriter(new OutputStreamWriter(connection.getOutputStream(), "UTF-8"));
writer.println("--" + boundary);
writer.println("Content-Disposition: form-data; name=\"paramToSend\"");
writer.println("Content-Type: text/plain; charset=UTF-8");
writer.println();
writer.println(paramToSend);
writer.println("--" + boundary);
writer.println("Content-Disposition: form-data; name=\"fileToUpload\"; filename=\"sample_signed.pdf\"");
writer.println("Content-Type: text/plain; charset=UTF-8");
writer.println();
BufferedReader reader = null;
try {
try {
reader = new BufferedReader(new InputStreamReader(new FileInputStream(fileToUpload), "UTF-8"));
} catch (UnsupportedEncodingException | FileNotFoundException e1) {
e1.printStackTrace();
}
try {
for (String line; (line = reader.readLine()) != null;) {
writer.println(line);
}
} catch (IOException e) {
e.printStackTrace();
}
} finally {
if (reader != null) try { reader.close(); } catch (IOException logOrIgnore) {}
}
writer.println("--" + boundary + "--");
} catch (UnsupportedEncodingException e2) {
e2.printStackTrace();
} catch (IOException e2) {
e2.printStackTrace();
} finally {
if (writer != null) writer.close();
}
// Connection is lazily executed whenever you request any status.
int responseCode = 0;
try {
responseCode = ((HttpURLConnection) connection).getResponseCode();
} catch (IOException e) {
e.printStackTrace();
}
System.out.println(responseCode);
Using only URLConnections.. This code isn't working properly.. It sends the file, but some content is lost and I don't know why.. My HTML sample works perfectly..
Can you help me?
Thank you for your attention,
Best Regards.
You are using a reader and writer combination, while you are transferring binary data. You should use streams when working with binary data.
So instead of a BufferedReader, use a normal InputStream buffered to read the file and do not wrap the output stream you get from the URLConnection.
Read bytes instead of strings in the read loop. So replace the part starting with BufferedReader until the finally just before the boundary write with:
OutputStream output = connection.getOutputStream();
InputStream fileIn = new FileInputStream(fileToUpload);
try {
byte[] buffer = new byte[4096];
int length;
while ((length = fileIn.read(buffer)) > 0) {
output.write(buffer, 0, length);
}
output.flush();
} finally {
if (fileIn != null) try { fileIn.close(); } catch (IOException logOrIgnore) {}
}
Solution: Another approach, using DataOutputStream.
FileInputStream fileInputStream = new FileInputStream(absolutePath);
URL url = new URL("http://localhost:50447/api/files/");
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("uploaded_file", filename);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\"" + fileName + "\"" + lineEnd);
dos.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
String serverResponseMessage = conn.getResponseMessage();
System.out.println(serverResponseMessage);
fileInputStream.close();
dos.flush();
dos.close();

How do I add REQUEST values to an HTTP POST method using multipart to upload a file to a PHP server in Android?

So, I'm trying to upload a file to my PHP server. I found some code online which works, but I also need to include values for things like user authentication and where on the server the file should be uploaded. I am relatively new to HTTP communication and the code I found below uses terms/code that I have never heard of before, multipart/form-data Content-Type and Content-Disposition specifically. So if someone could tell me how to include the values I need, provide a different method entirely, or just explain those 3 terms to me like I'm five, I'd greatly appreciate it. Here's my code:
public static void upload(String path, String section, Context c){
Log.i("path", path);
HttpURLConnection conn = null;
DataOutputStream dos = null;
DataInputStream inStream = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
String responseFromServer = "";
String urlString = c.getString(R.string.server) + "upload.php";
try {
// ------------------ CLIENT REQUEST
FileInputStream fileInputStream = new FileInputStream(new File(
path));
// open a URL connection to the Servlet
URL url = new URL(urlString);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
// Allow Inputs
conn.setDoInput(true);
// Allow Outputs
conn.setDoOutput(true);
// Don't use a cached copy.
conn.setUseCaches(false);
// Use a post method.
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\""
+ path + "\"" + lineEnd);
dos.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// close streams
Log.e("Debug", "File is written");
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
Log.e("Debug", "error: " + ex.getMessage(), ex);
} catch (IOException ioe) {
Log.e("Debug", "error: " + ioe.getMessage(), ioe);
}
// ------------------ read the SERVER RESPONSE
try {
inStream = new DataInputStream(conn.getInputStream());
String str;
while ((str = inStream.readLine()) != null) {
Log.e("Debug", "Server Response " + str);
}
inStream.close();
} catch (IOException ioex) {
Log.e("Debug", "error: " + ioex.getMessage(), ioex);
}
}
**Edit: Just to be a little more clear, I would like it to where in my PHP script I could access values like $_REQUEST['path'] (which might = '/documents/' or something) as well as the actual file with $_FILES['uploadedfile']
String urlToSendRequest = "https://example.net";
String targetDomain = "example.net";
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpHost targetHost = new HttpHost(targetDomain, 80, "http");
HttpPost httpPost = new HttpPost(urlToSendRequest);
// Make sure the server knows what kind of a response we will accept
// httpPost.addHeader("Accept", "text/xml");
// Also be sure to tell the server what kind of content we are sending
httpPost.addHeader("Content-Type", "application/xml");
StringEntity entity = new StringEntity("<input>test</input>", "UTF-8");
entity.setContentType("application/xml");
httpPost.setEntity(entity);
HttpResponse response = httpClient.execute(httpPost, context);
Reader r = new InputStreamReader(response.getEntity().getContent());

How to upload Bitmap Image from a android device?

Thank you in advance.
I'd like to upload some bitmap image from my android app.
but , I can't get it.
Could you recommend some solutions for it.
or collect my source code?
ByteArrayOutputStream bao = new ByteArrayOutputStream();
bitmap.compress(Bitmap.CompressFormat.JPEG, 90, bao);
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(
"http://example.com/imagestore/post");
MultipartEntity entity = new MultipartEntity( HttpMultipartMode.BROWSER_COMPATIBLE );
byte [] ba = bao.toByteArray();
try {
entity.addPart("img", new StringBody(new String(bao.toByteArray())));
httppost.setEntity(entity);
} catch (UnsupportedEncodingException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
// Execute HTTP Post Request
HttpResponse response = null;
try {
response = httpclient.execute(httppost);
} catch (ClientProtocolException e) {
}
use httpmime for uploading Image
try this
http://vikaskanani.wordpress.com/2011/01/11/android-upload-image-or-file-using-http-post-multi-part/
I found this solution really well created and 100% working even with amazon ec2, take a look into this link:
Uploading files to HTTP server using POST on Android (link deleted).
Compare to previous answer, this solution doesn't require to import huge library httpmime from Apache.
Copied text from original article:
This tutorial shows a simple way of uploading data (images, MP3s, text files etc.) to HTTP/PHP server using Android SDK.
It includes all the code needed to make the uploading work on the Android side, as well as a simple server side code in PHP to handle the uploading of the file and saving it. Moreover, it also gives you information on how to handle the basic autorization when uploading the file.
When testing it on emulator remember to add your test file to Android’s file system via DDMS or command line.
What we are going to do is set the appropriate content type of the request and include the byte array as the body of the post. The byte array will contain the contents of a file we want to send to the server.
Below you will find a useful code snippet that performs the uploading operation. The code includes also server response handling.
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
DataInputStream inputStream = null;
String pathToOurFile = "/data/file_to_send.mp3";
String urlServer = "http://192.168.1.1/handle_upload.php";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
try
{
FileInputStream fileInputStream = new FileInputStream(new File(pathToOurFile) );
URL url = new URL(urlServer);
connection = (HttpURLConnection) url.openConnection();
// Allow Inputs & Outputs.
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
// Set HTTP method to POST.
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);
outputStream = new DataOutputStream( connection.getOutputStream() );
outputStream.writeBytes(twoHyphens + boundary + lineEnd);
outputStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + pathToOurFile +"\"" + lineEnd);
outputStream.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// Read file
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0)
{
outputStream.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
outputStream.writeBytes(lineEnd);
outputStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
serverResponseCode = connection.getResponseCode();
serverResponseMessage = connection.getResponseMessage();
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception ex)
{
//Exception handling
}
If you need to authenticate your user with a username and password while uploading the file, the code snippet below shows how to add it. All you have to do is set the Authorization headers when the connection is created.
String usernamePassword = yourUsername + “:” + yourPassword;
String encodedUsernamePassword = Base64.encodeToString(usernamePassword.getBytes(), Base64.DEFAULT);
connection.setRequestProperty (“Authorization”, “Basic ” + encodedUsernamePassword);
Let’s say that a PHP script is responsible for receiving data on the server side. Sample of such a PHP script could look like this:
<?php
$target_path = "./";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path))
{
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded";
}
else
{
echo "There was an error uploading the file, please try again!";
}
?>;
Code was tested on Android 2.1 and 4.3. Remember to add permissions to your script on server side. Otherwise, the uploading won’t work.
chmod 777 uploadsfolder
Where uploadsfolder is the folder where the files are uploaded. If you plan to upload files bigger than default 2MB file size limit. You will have to modify the upload_max_filesize value in the php.ini file.

multipart file-upload post request from java

I'm trying to make a program that uploads a image to a webserver that accepts multipart file-uploads.
More specificly i want to make a http POST request to http://iqs.me that sends a file in the variable "pic".
I've made a lot of tries but i don't know if i've even been close. The hardest part seems to be to get a HttpURLConnection to make a request of the type POST. The response i get looks like it makes a GET.
(And i want to do this without any third party libs)
UPDATE: non-working code goes here (no errors but doesn't seem to do a POST):
HttpURLConnection conn = null;
BufferedReader br = null;
DataOutputStream dos = null;
DataInputStream inStream = null;
InputStream is = null;
OutputStream os = null;
boolean ret = false;
String StrMessage = "";
String exsistingFileName = "myScreenShot.png";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
String responseFromServer = "";
String urlString = "http://iqs.local.com/index.php";
try{
FileInputStream fileInputStream = new FileInputStream( new File(exsistingFileName) );
URL url = new URL(urlString);
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true);
conn.setDoOutput(true);
conn.setRequestMethod("POST");
conn.setUseCaches(false);
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);
dos = new DataOutputStream( conn.getOutputStream() );
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"pic\";" + " filename=\"" + exsistingFileName +"\"" + lineEnd);
dos.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0){
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
fileInputStream.close();
dos.flush();
dos.close();
}catch (MalformedURLException ex){
System.out.println("Error:"+ex);
}catch (IOException ioe){
System.out.println("Error:"+ioe);
}
try{
inStream = new DataInputStream ( conn.getInputStream() );
String str;
while (( str = inStream.readLine()) != null){
System.out.println(str);
}
inStream.close();
}catch (IOException ioex){
System.out.println("Error: "+ioex);
}
Two things:
Make sure you call setRequestMethod to set the HTTP request to be a POST. You should be warned that doing multipart POST requests by hand is difficult and error-prone.
If you're running on *NIX, the tool netcat is very useful for debugging this stuff. Run
netcat -l -p 3000
and point your program to port 3000; you'll see exactly what the program is sending (Control-C to close it afterwards).
I have used this and found it useful in multipart file upload
File f = new File(filePath);
PostMethod filePost = new PostMethod(url);
Part[] parts = { new FilePart("file", f) };
filePost.setRequestEntity(new MultipartRequestEntity(parts,
filePost.getParams()));
HttpClient client = new HttpClient();
status = client.executeMethod(filePost);

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