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• PHP ffmpeg output location

  18 janvier 2016, par Mick Jack

  My website allow users to upload video files to the server. Original file will be move to folder "Original" ffmpeg will compress the video and store the video in 480p folder. how to i setup ffmpeg so that it will store output of the compressed video file into original/480p ?

  Partial code of upload.php

  $target_dir = "Original/"; //where you want to upload the files to
  
  $target_file = $target_dir.basename($_FILES['file']['name']);
  
  $fileType = pathinfo($target_file, PATHINFO_EXTENSION);
  
  $newFileName = $target_dir.sha1(pathinfo(basename($_FILES['file']['name']), PATHINFO_FILENAME)).'480p'.'.'.$fileType;
  
  //$newFileName = $target_dir.sha1(pathinfo(basename($_FILES['file']['name']), PATHINFO_FILENAME)).'-'.time().'.'.$fileType;
  
  move_uploaded_file($_FILES['file']['tmp_name'], $newFileName);

  using this FFMPEG command

  shell_exec("C:\\ffmpeg\\bin\\ffmpeg.exe -y -i ".$newFileName." -c:v libx264 -s:v 854x480 -c:a copy \"480p\\{$newFileName}\" > logfile.txt 2>&1");

  i get the error
  480p\Original/ffbaf58f1231628f9ac2a583f038b51719006ec6480p.mp4 : No such file or directory

  i will like the output to be stored in original/480p/compressed-video-file

• Strange error with ffmpeg and unoconv in python script

  27 mai 2015, par Avery Ripoll Crocker

  I am creating a python script that can be used by other people to convert files. However when I use this script I keep getting a weird error when I run the subprocess for unoconv, this error is :

  Traceback (most recent call last):
  
    File "conversion.py", line 15, in <module>
  
      check_call(["unoconv", "-f", Fileextension, filename])
  
    File "/usr/lib/python2.7/subprocess.py", line 540, in check_call
  
      raise CalledProcessError(retcode, cmd)
  
  subprocess.CalledProcessError: Command '['unoconv', '-f', '.pdf', 'journal.doc']' returned non-zero exit status 1
  
  </module>

  I have looked through various resources for this, but the one answer I have been receiving is that I must have set up the unoconv line incorrectly. I have checked several times that the code is correct, and it is. The other peculiar thing is that my code for ffmpeg works. Does anybody have an idea as to why this happens.

  Here is my program :

  print "If at any point you wish to quit the program hit Ctrl + C"
  
  
  
  filetype = raw_input("What kind of file would you like to convert? Audio, Image, Video or Document: ")
  
  
  
  if filetype == "Document":
  
     path = raw_input("Please drag and drop the directory in which the file is stored into the terminal:")
  
     os.chdir(path[1:-2])
  
     filename = raw_input("Please enter the name of the file you would like to convert, including the file-type. e.g. test.txt, however please do make sure that the file-name does not have any spaces:")
  
     Fileextension = raw_input("What filetype would you like the program to convert your file to. E.g. .txt: ")
  
     from subprocess import check_call
  
     check_call(["unoconv", "-f", Fileextension, filename])
  
  
  
  elif filetype == "Audio":
  
     path = raw_input("Please drag and drop the directory in which the file is stored into the terminal:")
  
     os.chdir(path[1:-2])
  
     filename = raw_input("Please enter the name of the file you would like to convert, including the file-type. e.g. test.txt, however please do make sure that the file-name does not have any spaces:")
  
     Fileextension = raw_input("What filetype would you like the program to convert your file to. E.g. .mp3: ")
  
     body, ext = os.path.splitext("filename")
  
     from subprocess import check_call
  
     check_call(["ffmpeg" ,"-i", filename, body + Fileextension])
  
  
  
  elif filetype == "Video":
  
     path = raw_input("Please drag and drop the directory in which the file is stored into the terminal:")
  
     os.chdir(path[1:-2])
  
     filename = raw_input("Please enter the name of the file you would like to convert, including the file-type. e.g. test.txt, however please do make sure that the file-name does not have any spaces:")
  
     Fileextension = raw_input("What filetype would you like the program to convert your file to. E.g. .mp4: ")
  
     body, ext = os.path.splitext("filename")
  
     from subprocess import check_call
  
     check_call(["ffmpeg" ,"-i", filename, body + Fileextension])
  
  
  
  elif filetype == "Image":
  
     path = raw_input("Please drag and drop the directory in which the file is stored into the terminal:")
  
     os.chdir(path[1:-2])
  
     filename = raw_input("Please enter the name of the file you would like to convert, including the file-type. e.g. test.txt, however please do make sure that the file-name does not have any spaces:")
  
     Fileextension = raw_input("What filetype would you like the program to convert your file to. E.g. .Jpeg: ")
  
     body, ext = os.path.splitext("filename")
  
     from subprocess import check_call    
  
     check_call(["ffmpeg" ,"-i", filename, body + Fileextension])

• PHP upload video files to database

  13 janvier 2016, par Mick Jack

  I am working on a school project that let users upload video files to a server. Server will compress the video using ffmpeg and store the file in upload folder. Other users will be able to stream the uploaded videos.

  My question is how do i retrieve the video that ffmpeg generated and store the link in the database ?

  i am using this code but it only retrieve path of the original video.

  $filePath = dirname(__FILE__); 

  partial code of Upload.php

  $target_dir = "upload/"; //where you want to upload the files to
  
  $target_file = $target_dir.basename($_FILES['file']['name']);
  
  $fileType = pathinfo($target_file, PATHINFO_EXTENSION);
  
  $newFileName = $target_dir.sha1(pathinfo(basename($_FILES['file']['name']), PATHINFO_FILENAME)).'-'.time().'.'.$fileType;
  
  move_uploaded_file($_FILES['file']['tmp_name'], $newFileName);
  
  $unique_id = rand(1000000,9999999);
  
  
  
  shell_exec("C:\\ffmpeg\\bin\\ffmpeg.exe -i ".$newFileName." -vcodec libx264 -crf 20 \"upload\\{$newFileName}\" > logfile.txt 2>&1");
  
  
  
  /// save information into database
  
                  $username = "root";
  
                  $password = "";
  
                  $hostname = "localhost"; 
  
                  $dbname = "test_database";
  
  
  
                  //connect to the database
  
                  $dbc = mysqli_connect($hostname, $username, $password, $dbname) or die ("could not connect to the database");
  
  
  
                  //execute the SQL query and return records
  
                   $result = mysqli_query($dbc, "INSERT INTO `viewvideo` (`vID`, 'video_id`, `video_link`) VALUES ('', '".$unique_id."', '".$newFileName."')");
  
  
  
                  if(!$result){echo mysqli_error($dbc); }
  
  
  
                  echo $result;
  
  
  
              /*  
  
                  declare in the order variable
  
                  $result = mysqli_query($dbc, $sql); //order executes
  
                  if($result){
  
                      echo("<br />Input data is succeed");
  
                  } else{
  
                      echo("<br />Input data is fail");
  
                  }
  
              */  
  
  
  
                  //close the connection
  
                  mysqli_close($dbc);

  output