我已经将模拟器捕获的mp4视频上传到php服务器,但我无法使用任何播放器(窗口媒体播放器,vlc等)播放此上传的视频。
当我使用文件浏览器从模拟器中提取此vidoe时,它正在使用vlc媒体播放器。
protected String doInBackground(Void... params)
{
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
DataInputStream inputStream = null;
String urlServer = "http://10.0.2.2:80/maria/upload_file2.php";
String pathToOurFile="/storage/sdcard/DCIM/Camera/VID_20150225_152244.mp4";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "123*****sdf";
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=123*****sdf");
outputStream = new DataOutputStream( connection.getOutputStream() );
outputStream.writeBytes(twoHyphens + boundary + lineEnd);
outputStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + pathToOurFile );
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.d("ServerCode",""+serverResponseCode);
Log.d("serverResponseMessage",""+serverResponseMessage);
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception ex)
{
ex.printStackTrace();
}
return "Success";
}
服务器端的PHP代码。
$targetfolder = "images/";
$targetfolder = $targetfolder . basename( $_FILES['uploadedfile']['name']) ;
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $targetfolder))
{
echo "The file ". basename( $_FILES['uploadedfile']['name']). " is uploaded";
}
else {
echo "Problem uploading file";
}
我已经通过从html表单上传视频测试了PHP代码,这个视频播放正常,我觉得android代码有问题。有人可以帮助我在代码中找到问题。
答案 0 :(得分:0)
在php服务器上传视频时,请在您的数据库中保留每个视频名称及其直接mp4网址。例如,在您的PHP代码中,您将视频上传到服务器的图像文件夹中。如果您的视频名称为“myvideo”,则其网址为http://your-domain-name.com/images/myvideo.mp4。将这些链接保存在您的数据库中,并使Web服务获取所有视频链接。从android调用该Web服务并使用android videoview播放该视频。这是视频的xml。
<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:orientation="vertical" >
<VideoView
android:id="@+id/videoView"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:layout_alignParentBottom="true"
android:layout_alignParentLeft="true"
android:layout_alignParentRight="true"
android:layout_alignParentTop="true" />
</RelativeLayout>
这里是播放视频的代码,不需要asynctask在android中播放视频。 videoview将流式传输。
VideoView videoView =(VideoView)findViewById(R.id.videoView);
MediaController mediaController= new MediaController(this);
mediaController.setAnchorView(videoView);
Uri uri=Uri.parse(" http://your-domain-name.com/images/myvideo.mp4");
videoView.setMediaController(mediaController);
videoView.setVideoURI(uri);
videoView.requestFocus();
videoView.setOnPreparedListener(new MediaPlayer.OnPreparedListener() {
@Override
public void onPrepared(MediaPlayer mp) {
// TODO Auto-generated method stub
videoView.start();
}
});