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File upload in Java from Java client using HTTP

My client code is in java. It uploads file to different server based on request. These request go between countries like "India to US", "US to UK" etc. Also, sometimes files are as large as 2 GB. My current code is ineffective as it is slow in uploading file and requires you to give 1 GB heap space even to upload just 60 MB file.
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
//System.out.println("Uploading the file");
String responseFromServer = "";
String urlString = "http://stcss.us.com:8888/codesign/UploadServlet";
try {
//------------------ CLIENT REQUEST
FileInputStream fileInputStream =
new FileInputStream(new File(exsistingFileName));
int file_size = fileInputStream.available();
if (file_size > 1000 * 1024 * 1024) {
System.out.println("File Size Error\n Max Size allowed is 500MB");
System.exit(1);
}
// open a URL connection to the Jsp
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);
//conn.setRequestProperty("Accept", "application/octet-stream");
conn.setRequestProperty("password", password);
conn.setRequestProperty("signingParameters", signingParameters);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"upload\";" +
" filename=\"" + exsistingFileName + "\"" +
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) {
// System.out.println ("file size is"+file_size+"avaiable is"+bytesAvailable);
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
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
System.out.println("From CLIENT CLIENT REQUEST Malformed:" + ex);
System.exit(0);
} catch (IOException ioe) {
System.out.println("From CLIENT CLIENT REQUEST:" + ioe);
System.exit(0);
}
What are the ways I can change this code to make upload faster and also get rid of heap space for considerably smaller file? Is something neeeded to be changed on server side? I am using apache-commons-fileupload on the server side.

trying to upload a image to my server from android

am using this code to upload an image to my web server:
public void send (View view)
{
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
DataInputStream inputStream = null;
String pathToOurFile = "sdcard/yo.jpg";
String urlServer = "http://192.168.1.4/uplaod.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);
// Enable POST method
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)
int serverResponseCode = connection.getResponseCode();
String serverResponseMessage = connection.getResponseMessage();
fileInputStream.close();
outputStream.flush();
outputStream.close();
Log.i("ODPOWIEDZ", serverResponseMessage);
Context context = getApplicationContext();
Toast.makeText(context, "its here 1 ", Toast.LENGTH_LONG).show();
} catch (Exception ex) {
Context context = getApplicationContext();
Toast.makeText(context, "its in catsh", Toast.LENGTH_LONG).show();
Log.i("WYJATEK", ex.getMessage());
}
}
and i get this message from my LogCat where is the problem here ??
59-124/system_process D/SntpClient﹕ request time failed: java.net.SocketException: Address family not supported by protocol
this is my php file
<?php
$target_path = "uploadedfile/";
$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!";
}
?>

The SNTP error relates to the time protocol, which, incidentally has nothing to do with your app. It looks like it's just the emulator trying to fetch the latest timestamp off an ntp server. Just discard it.
The issue is still something else. You did notice that there might be a typo in the name of the php script? In your code you refer to it as uplaod.php.
Also, make sure you have enabled internet access in your manifest file:
<uses-permission android:name="android.permission.INTERNET"></uses-permission>

Error upload image in Android app -- Api Foursquare

I am developing an Android application in which users can upload photos of venues you visit. I'm using the endpoints and have begun testing https://api.foursquare.com/v2/photos/PHOTO_ID through Apigee console. The problem I have is that if I send an image via the console returns me an error 400 problems with mime type as you can in this still image is correct, I think.
https://docs.google.com/file/d/0B9uUMZ3ZVbl_bG5rOUNhM01wb2M/edit?usp=sharing (console error)
I have also tried uploading through the application I am developing and I returned the same error. The execute method in the app is as follows:
TokenStore tokenStore=new TokenStore(getApplicationContext());
String token=tokenStore.getSavedToken();
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
DataInputStream inputStream = null;
String pathToOurFile = params[0];
String urlServer = "https://api.foursquare.com/v2/photos/add?public=1&venueId=5252da418bbd79f3aaa70ae6&oauth_token="+token;
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);
// Enable POST method
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type", "image/jpeg");
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("Content-Type: image/jpeg");
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)
int serverResponseCode = connection.getResponseCode();
String serverResponseMessage = connection.getResponseMessage();
Log.i("serverResponse", "serverresponse "+ serverResponseMessage);
Log.i("serverResponseCode", "serverResponseCode "+ serverResponseCode);
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception ex) {
//Exception handling
}
In my tests online server, upload the image class works fine so the problem must be with some kind of data that does not put right for the Foursquare API. Anyone have experience with this? thanks

POSTing an image to PHP file

I'm trying to POST an image to a PHP file from my android app and am wondering what format (File, Fileoutputstream etc) I have to post it in for it to be recognized as a file and refer to it with $_FILE['filename'] in my php script.
Thanks :)
EDIT:
Sorry I may not have been clear, I'm not looking for the PHP script, I already have that finished accepting the $_FILE['sample'] and doing what I need with it, I'm just not sure the file TYPE that I have to post to the php file (IN JAVA) in order for php to 'see' it as $_FILE
FYI: I am using the loopj asynchronous http request library.
public void add_image_android(final Bitmap image, String party_id, String guest_id)
{
String url = "http://www.mysite.com/urltopost";
/* not sure what to set fOut to for the bitmap to be passed as file */
RequestParams params = new RequestParams();
params.put("file", fOut);
params.put("guest_id", guest_id);
params.put("party_id", party_id);
client.post(url, params, new JsonHttpResponseHandler()
{
#Override
public void onSuccess(JSONObject response)
{
((ResponseListener)_mainContext).add_image_android_response(response.toString());
return;
}
#Override
public void onFailure(Throwable e)
{
fireToast("api error:"+e);
Log.d("api error:",e.toString());
}
});
}

Try the Below Code which will upload the Image and Give the Link.
<?php
$uploaddir = 'images/';
$ran = rand () ;
$file = basename($_FILES['userfile']['name']);
$uploadfile = $uploaddir .$ran.$file;
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
echo "http://www.domain.com/folder/{$uploadfile}";
}
?>

This worked for me: (very old code, hope it helps...)
ReturnObject returnObject = new ReturnObject();
HttpURLConnection conn = null;
DataOutputStream dos = null;
BufferedReader inStream = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
String urlString = "your url";
try{
FileInputStream fileInputStream = new FileInputStream(photoFile);
URL url = new URL(urlString);
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true);
conn.setDoOutput(true);
conn.setUseCaches(false);
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=\"image\";"
+ " filename=\"" + photoFile.getAbsolutePath() +"\"" + 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){
ex.printStackTrace();
}catch (IOException ioe){
ioe.printStackTrace();
}
On the server i found this:
$source = $_FILES['image']['tmp_name'];
move_uploaded_file($source, $target)
Not sure what this "tmp_name" is...

Uploading an image to a server in android

I my app I am trying to send an image over a PHP server. The problem is the PHP programmers say that it should be sent as an data. Following is the code which I am currently using, please help me to convert the image to a data and get a response from the server.
InputStream is;
private int serverResponseCode;
private String serverResponseMessage;
#Override
public void onCreate(Bundle icicle)
{
super.onCreate(icicle);
setContentView(R.layout.main);
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
DataInputStream inputStream = null;
String pathToOurFile = "/sdcard/siva.PNG";
Log.e("pathToOurFile",""+pathToOurFile);
String urlServer = "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX/upload.php";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
Log.e("URL Server",""+urlServer);
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
Log.e("maxBufferSize",""+maxBufferSize);
try
{
FileInputStream fileInputStream = new FileInputStream(new File(pathToOurFile) );
Log.e("FIS",""+fileInputStream);
URL url = new URL(urlServer);
connection = (HttpURLConnection) url.openConnection();
// Allow Inputs & Outputs
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
// Enable POST method
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)
Log.e("con",String.valueOf(connection.getDoOutput()));
serverResponseCode = connection.getResponseCode();
serverResponseMessage = connection.getResponseMessage();
Log.e("response",""+serverResponseCode);
Log.e("serverResponseMessage",""+serverResponseMessage);
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception ex)
{
Log.e("Exception Handling",""+ex);
}
}

Atlast i found the answer, i did a small mistake....In the above code i have changed only one word in the following line
outputStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + pathToOurFile +"\"" + lineEnd);
In the above line the main part is the word "uploadedfile". This word must be specified by the php programmer or else the file which we are sending will not be replaced.
Please refer the here