I want to copy all the video files on my server & save it in the web contents folder of a web service , so that it can be visible to all later on !
How should i proceeed ?
I want to copy all the video files on my server & save it in the web contents folder of a web service , so that it can be visible to all later on !
How should i proceeed ?
Basically you need two sides. The first one is on Android. You have to send (Do it in an ASyncTask e.x.) the data to your webservice. Here I made a little method for you, which sends a file and some additional POST values to an URL:
private boolean handleFile(String filePath, String mimeType) {
HttpURLConnection connection = null;
DataOutputStream outStream = null;
DataInputStream inStream = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
String urlString = "http://your.domain.com/webservice.php";
try {
FileInputStream fileInputStream = null;
try {
fileInputStream = new FileInputStream(new File(filePath));
} catch(FileNotFoundException e) { }
URL url = new URL(urlString);
connection = (HttpURLConnection) url.openConnection();
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
SharedPreferences prefs = PreferenceManager.getDefaultSharedPreferences(getBaseContext());
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);
outStream = new DataOutputStream(connection.getOutputStream());
// Some POST values
outStream.writeBytes(addParam("additional_param", "some value");
outStream.writeBytes(addParam("additional_param 2", "some other value");
// The file with the name "uploadedfile"
outStream.writeBytes(twoHyphens + boundary + lineEnd);
outStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + filePath +"\"" + lineEnd + "Content-Type: " + mimeType + lineEnd + "Content-Transfer-Encoding: binary" + lineEnd);
outStream.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
outStream.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
outStream.writeBytes(lineEnd);
outStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
fileInputStream.close();
outStream.flush();
outStream.close();
} catch (MalformedURLException e) {
Log.e("APP", "MalformedURLException while sending\n" + e.getMessage());
} catch (IOException e) {
Log.e("APP", "IOException while sending\n" + e.getMessage());
}
// This part checks the response of the server. If its "UPLOAD OK" the method returns true
try {
inStream = new DataInputStream( connection.getInputStream() );
String str;
while (( str = inStream.readLine()) != null) {
if(str=="UPLOAD OK") {
return true;
} else {
return false;
}
}
inStream.close();
} catch (IOException e){
Log.e("APP", "IOException while sending\n" + e.getMessage());
}
return false;
}
Now, we have got the other side: http://your.domain.com/webservice.php Your server. It needs some logic, for example in PHP, to handle the sent POST request. Something like this would work:
<?php
$target_path = "videos/";
$target_path = $target_path.'somename.3gp';
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
exit("UPLOAD OK");
} else {
exit("UPLOAD NOT OK");
}
?>