[英]How To Upload Video in Android to Server via Webservice?
I have captured a video and i want to upload to server via webservice call.I've tried with volley library ie, how we upload image i have tried the same for video but its not uploading.Please suggest me some good libraries for uploading videos in android. 我已经捕获了一个视频,我想通过webservice调用上传到服务器。我已经尝试了凌空库,即,我们如何上传图像,我已经尝试了相同的视频但没有上传。请建议我一些不错的库来上传视频在android中。
Thanks, Rojesh 谢谢罗吉什
Use below function to upload video to server... 使用以下功能将视频上传到服务器...
private void doFileUpload(){
HttpURLConnection conn = null;
DataOutputStream dos = null;
DataInputStream inStream = null;
String lineEnd = "rn";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
String responseFromServer = "";
String urlString = "http://your_website.com/upload_audio_test/upload_audio.php";
try
{
//------------------ CLIENT REQUEST
FileInputStream fileInputStream = new FileInputStream(new File(selectedPath) );
// 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="" + selectedPath + """ + 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);
}
}
PHP web-service. PHP Web服务。
<?php
// Where the file is going to be placed
$target_path = "uploads/";
/* Add the original filename to our target path.
Result is "uploads/filename.extension" */
$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!";
echo "filename: " . basename( $_FILES['uploadedfile']['name']);
echo "target_path: " .$target_path;
}
?>
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