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• Getting a single mp4 file from dash and a mpd file [closed]

  15 août 2023, par Govind Sharma

  I want to ask that I have downloaded data which contains audio and video data separately like 1.mp4, 2.mp4 etc in both audio and video folders. enter image description here
I also have it's .mpd file. How can I play this video ? or is there any way I can combine both videos and audio files in a single one. Single file for e.g. 1.mp4 video or audio nothing seem to play.

  


  I trued using ffmpeg but it gave a blurr video

  


• To convert from avi file to mp4 file by ffmpeg command in php

  12 septembre 2018, par user27240

  The ffmpeg command in php below creates mp4 file from avi file without deleting the original avi file meaning that,for example,
  there are two files, 1221222.avi and 1221222.mp4 exist in the directry after all.

  (’for i in /xxxxx/xxxxxx/*.avi ; do ffmpeg -i "$i" -frames:v 1 "/xxxxx/xxxxxx/$(basename "$i" .avi).mp4" ; done’)

  I’d like to delete this 1221222.avi file right after the command created 1221222.mp4 file.

  Is there any command that converts the file from avi to mp4 by replacing it ? Or do I have to add delete command after the command above ?

  I would appreciate if anyone could help me out for the command.

  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

  Even though the php script below can upload avi file from local drive to application I’m currently developping, the avi will not be played at html.
  So, I’m trying to convert from the avi file to mp4 file which is working fine by using ffmpeg command in php. This ffmpeg command creates mp4, but the html still do not play it. Is it because the original avi file still exists in the directory ?

  //php
  
  case "video/x-msvideo":
  
  
  
  if($header){
  
      header("Content-Type: video/x-msvideo");
  
  
  
   $dst_im = copy($path, $dst_file);
  
   return "";
  
   }else{ 
  
  
  
   $dst_file = $dst_file . ".avi";
  
   shell_exec('for i in  /xxxxxx/xxxxxx/*.avi; do ffmpeg -i "$i" - frames:v 1 "/xxxxxx/xxxxxx/$(basename "$i"  .avi).mp4"; done');      
  
  
  
   $dst_im = copy($path, $dst_file);
  
   }
  
   unlink($dst_im);
  
   break; 
  
  
  
  
  
  //html
  
  <video width="500" height="250" poster="video.jpg" controls="controls" preload="auto" autoplay="autoplay">
  
  <source type="video/mp4" src="xxxxxxxxxxxxxxxxxxxxx.mp4"></source>
  
  </video>

• ffmpeg is always giving me No such File TEST8.mp4 but file is there and this command is getting executed directy in console

  13 juin 2013, par finch986

  ffmpeg is always giving me No such File TEST8.mp4 but file is there and this command is getting executed directy in console
  ffmpeg is always giving me No such File TEST8.mp4 but file is there and this command is getting executed directy in console

  String exe = "chmod 777 /data/data/com.demo.videocomparedemo/ffmpeg"/*+" -i TEST8.mp4 -acodec copy -vf \"transpose=1\" TEST8-1.mp4"*/;
  
  Process process = runtime.exec(exe,null,null);
  
  process.waitFor();
  
  process.destroy();
  
  
  
  String src="/mnt/sdcard/Swingpro/TEST8.mp4";
  
  String exe1 = "chmod 777 "+src/*+" -i TEST8.mp4 -acodec copy -vf \"transpose=1\" TEST8-1.mp4"*/;
  
  Process process1 = runtime.exec(exe1,null,null);
  
  process1.waitFor();
  
  process1.destroy();
  
  
  
  File file = new File("/mnt/sdcard/Swingpro", "TEST9.mp4");
  
  boolean retVal = false;
  
  if(file.exists()){
  
      if(file.isDirectory()){
  
          retVal = file.delete();
  
      }
  
  }
  
  file.createNewFile();
  
  
  
  String src1="/mnt/sdcard/Swingpro/TEST9.mp4";
  
  String exe2 = "chmod 777 "+src1;
  
  Process process2 = runtime.exec(exe2,null,null);
  
  process2.waitFor();
  
  process2.destroy();
  
  
  
  
  
  File dir = new File("/mnt/sdcard/SwingPro");
  
  
  
  List<string> args = new ArrayList<string>();
  
  args.add ("/data/data/com.demo.videocomparedemo/ffmpeg"); // command name
  
  args.add ("-i "+src); // optional args added as separate list items
  
  args.add("-acodec copy ");
  
  args.add("-vf \"transpose=1\" "+src1);
  
  
  
  
  
  ProcessBuilder pb = new ProcessBuilder (args);
  
  Process p = pb.start();
  
  p.waitFor();
  
  </string></string>