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• ProcessBuilder is not called when trying to start a process

  15 juin 2022, par xnok

  I am trying to understand more about the ffmpeg usage in JavaCV for android studio and for said task I am trying to use ProcessBuilder. I tried writting a simple program to debug the pb.start(); Although, I am not getting a response. What I did was to start a default/empty activity and pasted the following program :

  


  package com.example.myapplication;

import androidx.annotation.RequiresApi;
import androidx.appcompat.app.AppCompatActivity;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;

import org.bytedeco.javacpp.Loader;

import android.os.Build;
import android.os.Bundle;
import android.util.Log;

public class MainActivity extends AppCompatActivity {
    static final int cols = 192;
    static final int rows = 108;
    static final String ffmpeg = Loader.load(org.bytedeco.ffmpeg.ffmpeg.class);
    static final String rtmp_url = "test.flv";
    static final String[] command = {ffmpeg,
            "-y",
            "-f", "rawvideo",
            "-vcodec", "rawvideo",
            "-pix_fmt", "bgr24",
            "-s", (Integer.toString(cols) + "x" + Integer.toString(rows)),
            "-r", "10",
            "-i", "pipe:",
            "-c:v", "libx264",
            "-pix_fmt", "yuv420p",
            "-preset", "ultrafast",
            "-f", "flv",
            rtmp_url};
    @RequiresApi(api = Build.VERSION_CODES.O)
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        new Thread(t1).start();

    }
    private static Runnable t1 = () -> {
        Log.e("TAG", "void OnCreate called successfully!");
        ProcessBuilder pb = new ProcessBuilder(command).redirectErrorStream(true);
        pb.redirectErrorStream(true);
        try {
            Process process = pb.start();
            BufferedReader reader = new BufferedReader(new InputStreamReader(process.getInputStream()));
            OutputStream writer = process.getOutputStream();
            Log.e("TAG", "Something good happened here");
        } catch (IOException e) {
            e.printStackTrace();
            Log.e("TAG", "Nothing good happened here");
        }
    };


}


  


  My current problem is that I can't seem to start properly the processBuilder process via pb.start() ;

  


  I get the following logs from the logcat panel :

  


  2022-06-14 17:24:46.328 13371-13371/com.example.myapplication E/TAG: void OnCreate called successfully!
2022-06-14 17:24:46.333 13371-13371/com.example.myapplication E/TAG: Nothing good happened here


  


  I'd like to understand why is it skipping the try/catch block and not starting the process ?

  


  EDIT : I made some changes as per @g00se's suggestions and I got the following stack trace from the code above :

  


  2022-06-15 00:32:26.700 29787-29787/? E/USNET: USNET: appName: com.example.myapplication
2022-06-15 00:32:29.328 29787-29828/com.example.myapplication E/TAG: void OnCreate called successfully!
2022-06-15 00:32:29.330 29787-29828/com.example.myapplication E/AndroidRuntime: FATAL EXCEPTION: Thread-4
    Process: com.example.myapplication, PID: 29787
    java.lang.NullPointerException
        at java.lang.ProcessBuilder.start(ProcessBuilder.java:1012)
        at com.example.myapplication.MainActivity.lambda$static$0(MainActivity.java:48)
        at com.example.myapplication.MainActivity$$ExternalSyntheticLambda0.run(Unknown Source:0)
        at java.lang.Thread.run(Thread.java:920)


  


• How can I convert a mp4 file to webm using java only ?

  31 octobre 2023, par mir-shakir

  I am trying to convert a file from mp4 to webm. I am trying to use the JAVE wrapper of FFmpeg. I am getting the error.
Here is my code :

  


  private String ConvertVideo(URL url) {
    File target =null;
    MultimediaObject multimediaObject = new MultimediaObject(url);
    try {
        target = File.createTempFile("target", ".webm");

    } catch (IOException e1) {
        e1.printStackTrace();
    }
    AudioAttributes audio = new AudioAttributes();
    audio.setCodec(AudioAttributes.DIRECT_STREAM_COPY);
    audio.setBitRate(new Integer(128000));
    audio.setSamplingRate(new Integer(44100));
    audio.setChannels(new Integer(2));
    VideoAttributes video = new VideoAttributes();
    video.setBitRate(new Integer(160000));
    video.setFrameRate(new Integer(15));
    video.setCodec("libvpx-vp9");
    video.setCodec(VideoAttributes.DIRECT_STREAM_COPY);
    EncodingAttributes attrs = new EncodingAttributes();
    attrs.setFormat("webm");
    attrs.setAudioAttributes(audio);
    attrs.setVideoAttributes(video);
    
    try {
          Encoder encoder = new Encoder();  
          encoder.encode(multimediaObject, target, attrs);
        } catch (Exception e) {  
           e.printStackTrace();
        }
    
    return "success";
}


  


  I am getting the below error :

  


  2022-Jun-13 11:12:55 AM [qtp1914526580-175] ERROR ws.schild.jave.Encoder - Process exit code: 1  to target2636257785060285182.webm
ws.schild.jave.EncoderException: Exit code of ffmpeg encoding run is 1


  


  What am I doing wrong here. Is there any other way around it ? I want to do it only using java.

  


• VLC RTSP HTML5 transcoding

  30 mai 2022, par Pierogi

  I'm trying to get audio streaming on an HTML page from an RTSP server.

  


  The RTSP server is the rtsp-simple-server running a command line below.
  
./rtsp-simple-server rtsp-simple-server.yml.
  
The configure file is the default.

  


  The stream player is FFmpeg running a command line below.
  
ffmpeg -re -stream_loop -1 -i myaudio.mp3 -c copy -f rtsp -rtsp_transport tcp rtsp://localhost:8554/mystream

  


  The console log at the time the rtsp-simple-server and the ffmpeg are started is below.

  


  2022/05/29 19:06:38 INF rtsp-simple-server v0.18.4
2022/05/29 19:06:38 INF [RTSP] listener opened on :8554 (TCP), :8000 (UDP/RTP), :8001 (UDP/RTCP)
2022/05/29 19:06:38 INF [RTMP] listener opened on :1935
2022/05/29 19:06:38 INF [HLS] listener opened on :8888
2022/05/29 19:09:16 INF [RTSP] [conn [::1]:62737] opened
2022/05/29 19:09:16 INF [RTSP] [session 271690815] created by [::1]:62737
2022/05/29 19:09:16 INF [RTSP] [session 271690815] is publishing to path 'mystream', 1 track with TCP


  


  And the time the rtsp path(rtsp ://localhost:8554/mystream) is opened by VLC, the contents can be played properly. The additional console log at the time is below.

  


  2022/05/29 19:13:19 INF [RTSP] [conn 127.0.0.1:62780] opened
2022/05/29 19:13:19 INF [RTSP] [session 734209460] created by 127.0.0.1:62780
2022/05/29 19:13:19 INF [RTSP] [session 734209460] is reading from path 'mystream', 1 track with UDP
2022/05/29 19:13:29 INF [RTSP] [session 734209460] destroyed (teared down by 127.0.0.1:62780)
2022/05/29 19:13:29 INF [RTSP] [conn 127.0.0.1:62780] closed (EOF)
2022/05/29 19:13:29 INF [RTSP] [conn 127.0.0.1:62781] opened
2022/05/29 19:13:29 INF [RTSP] [session 445756113] created by 127.0.0.1:62781
2022/05/29 19:13:29 INF [RTSP] [session 445756113] is reading from path 'mystream', 1 track with TCP


  


  However, I open the rtsp streaming from the VLC's "Network" tab like below,


  


  and configure the "Stream output" like below,


  


  and I tried to get this streaming from an HTML page like below,

  


  &#xA;&#xA;  &#xA;    &#xA;    &#xA;  &#xA;  &#xA;    <h1>transcode test</h1>&#xA;    <audio src="http://localhost:9999/mystream" autoplay="autoplay"></audio>&#xA;  &#xA;&#xA;

  


  the browser console displays Failed to load resource: the server responded with a status of 404 (Not found). I've already tried other ports(etc. 8080).

  


  So, how can I get the rtsp stream from the RTSP server on an HTML page.
Any idea ?

  


  My environment.

  


  

  • Browser : Microsoft edge

  


  • OS : MacOS 11.6.5

  


  • rtsp-simple-server : 0.18.4

  


  • FFmpeg : 5.0.1

  


  • VLC : 3.0.17.3