i'm trying to extract a single frame from a video file using the following php code:
$cmd = 'ffmpeg -i "d:\webs\beta\test\sample2.mp4" -vframes 1 -s 146x82 -f image2 "d:\webs\beta\test.jpg"';
exec($cmd, $rc);
the problem is that i'm getting an 500 internal server error the first time i'm trying to execute the script, but when reloading it works.
so it means when reloading: works / doesn't work / works ..
any ideas what could be wrong?
Try this script.
<?php
$ffmpeg = "/full/path/to/ffmpeg";
$videoFile = "/full/path/to/video.mp4";
$imgOut = "/full/path/to/frame.jpg";
$second = 0;
$cmd = $ffmpeg." -i \"".$videoFile."\" -an -ss ".$second.".001 -y -f mjpeg \"".$imgOut."\" 2>&1";
$feedback = `$cmd`;
?>
i had the same problem exactly.
using proc_open instead of exec and its variants fixed it.
Roey
Related
I am currently trying to convert various video files using FFmpeg with the help of my server (Windows Server 2016). PHP is supposed to initiate the process.
Whenever I start FFmpeg via CMD, everything works fine and the video file converts perfectly. But when I trigger the same process with PHP, FFmpeg aborts the conversion after a while and exits by itself.
I neither get an error in the log nor can I recognize system errors from Windows Server.
It doesn't matter if I run the command through shell_exec, exec, or system.
For converting I actually use this code:
ffmpeg -loglevel error -i $video -vf scale=1920:1080 -crf 10 -c:v libx264 -preset veryfast -threads 2 $output > NUL 2>&1 < NUL
This is my full php code:
ignore_user_abort(true);
set_time_limit(0);
error_reporting( E_ALL | E_STRICT );
$path = 'C:/inetpub/vhosts/confident-tharp.xx-xx-xxx-xxx.plesk.page/httpdocs';
$ffmpeg = $path . '/ffmpeg/ffmpeg.exe';
$video = $path . '/convert/myoldfile.mp4';
$output = $path . '/convertedfile.mp4';
$dimension = 'scale=1920:1080';
$command = "ffmpeg -loglevel error -i $video -vf scale=1920:1080 -crf 10 -c:v libx264 -preset veryfast -threads 2 $output > NUL 2>&1 < NUL";
shell_exec( $command );
I'm currently using FFmpeg 5, but there was no change with FFmpeg 4 either. The use of other codecs or file formats (e.g. .avi) also has no influence on the behavior of FFmpeg. Previously I tried to use the php-ffmpeg library. Again, I get the same "error".
Hope you guys can help.
I have a file located at html://www.example.com/wp-content/music.mp3
I've tested and confirmed ffmpeg is installed and have run
exec("ffmpeg -help",$output);
I successfully get an output. Now i want to start converting but i cannot locate the file above. I've tried
exec("ffmpeg -i html://www.example.com/wp-content/music.mp3",$output);
exec("ffmpeg -i home/mywebsite/public_html/wp-content/music.mp3",$output);
I get no output for either. ffmpeg is located in /usr/bin/ffmpeg.
How do i solve?
Redirect the error output (stderr) to response (stdout):
<?php
$command = "ffmpeg -i https://www.example/a.mp3 2>&1";
$output = shell_exec($command);
echo $output;
Have been trying to execute FFMPEG using a script I have uploaded to my domain
<?php
$output = array();
$result = -1;
exec('../../../../../../usr/bin/ffmpeg -ab 320k -i source.wav dest320.mp3', $output, $result);
var_dump($output, $result);
?>
The example code says the program should not be returning -1 unless there is an error but I have pointed to the exact path that FFMPEG is stored in.....
When I call 'ffmpeg -ab 320k -i source.wav dest320.mp3' from CentOS it works...
Am lost and have spent hte last few hours trying to work it out.
Thanks
CP
Any time you have an exec that doesn't work in php, you should switch to passthru for debugging.
passthru('../../../../../../usr/bin/ffmpeg -ab 320k -i source.wav dest320.mp3 1 2>&1');
Appending 1 2>&1 to the end, pipes your stderr to stout, and returns any errors you have while running your exec.
It is probably safer to use passthru('/usr/bin/env ffmpeg -ab 320k -i source.wav dest320.mp3 1 2>&1'); which will get the correct path for ffmpeg without the need to specify and absolute/relative path.
Also worth noting that if you are using Ubuntu since 12.04, the ffmpeg command has been renamed to avconv.
i have a simple site on which people upload videos, so i want to generate a simple thumbnail from an uploaded video. i have tried every trick and way to do this from a number of websites but i am failing to make the command run without problems.
$video = $_FILES['vpopupdropin']["tmp_name"];
$ffmpeg = "C:\\Ffmpeg\\ffmpeg-20130605-git-3289670-win64-static\\bin";
$image = "manu.jpg";
$second = 12;
$size = "150x90";
$command = "$ffmpeg -i $video -an -ss $second -s $size -vcodec mjpeg $image";
echo $command;
shell_exec($command);
if(shell_exec($command)){
echo 'okay';
echo '<img src="'.$image.'"/ >';
}
else{
echo ' Problem';
}
i Echoed the the command from PHP and this is what i got:
C:\Ffmpeg\ffmpeg-20130605-git-3289670-win64-static\bin -i C:\xampp\tmp\php27F1.tmp -an -ss 12 -s 150x90 -vcodec mjpeg manu.jpg Problem
so i took the Command above and entered it in Cmd and got this error
[image2 # 00000000000000003d87580] Could not open file : manu.jpg
av_interleaved_write_frame(): Input/output error. the uploaded file transfers well to where iam saving it and plays well on the site meaning the file is not corrupt. but the thumbnail command seems to fail, i have even checked the other questions on this site but i seem to fail to get the right solution. the paths in the Command are correct and i have verified that at least
You did not give ffmpeg a name :-) So you tried to execute a \\bin folder !
$ffmpeg = "C:\\Ffmpeg\\ffmpeg-20130605-git-3289670-win64-static\\bin";
you forget ffmpeg.exe
$ffmpeg = "C:\\Ffmpeg\\ffmpeg-20130605-git-3289670-win64-static\\bin\\ffmpeg";
I do it for a .avi with following command
ffmpeg -i Echo2012.avi -r 1 -s 1024x576 -f image2 -vframes 1 foo-001.jpg
Don't execute your command twice !
$command = "$ffmpeg -i $video -an -ss $second -s $size -vcodec mjpeg $image";
echo $command;
shell_exec($command);
if(shell_exec($command)){
EDIT :
your command string :
ffmpeg -i upload.tmp -an -ss 12 -s 150x90 -vcodec mjpeg manu.jpg
-vcodec codec (output) : Set the video output codec. It's a switch for a output video. You want as output an image.
-an : You can disable Audio stream. You don't need Audio for an image.
-ss : position (input/output) When used as an input option (before -i), seeks in this input file to position.
my command string :
ffmpeg -i Echo2012.avi -r 1 -s 1024x576 -f image2 -vframes 1 foo-001.jpg
-r : fps (input/output,per-stream) . Set frame rate (Hz value, fraction or abbreviation).
As an input option, ignore any timestamps stored in the file and instead generate timestamps assuming constant frame rate fps.
-f image2 : Force output file format image2. The format is normally auto detected guessed from the file extension for output files.
-vframes number (output) : Set the number of video frames to record.
I have read this but didnt help too much.
I have a folder called videos and another folder called thumbnails. I have many mp4 videos in video folder and want to catch thumbnails at 4th second to the thumbnails folder using ffmpeg and php.
I am using Wamp server 2.2 on windows whit php 5.3.8 and Apache 2.2.21
I downloaded ffmpeg from FFmpeg Windows Builds section of ffmpeg download page and the static 32 build from this link.
I extracted the 7z file to my website root
here is my php code:
$ffmpeg = "includes/ffmpeg/bin/ffmpeg";
foreach(glob('files/videos/*.mp4') as $pathname){
$filename = substr($pathname,13,strripos($pathname,'.mp4')-13);
$thumbnail = 'files/thumbnails/'.$filename.'.jpg';
exec("ffmpeg -i $pathname -an -y -f mjpeg -ss 00:00:04 -vframes 1 $thumbnail");
}
but nothing happens and the thumbnails folder is always empty!
- How can I find out is ffmpeg installed on my server or not?
- How can I get my script to work?
Please help
try this:
$ffmpeg = "c:/wamp/www/includes/ffmpeg/bin/ffmpeg";
$videos = "c:/wamp/www/files/videos/*.mp4";
$ouput_path = "c:/wamp/www/files/thumbnails/";
foreach(glob($videos) as $v_file){
$fname = basename($v_file, ".mp4");
$thmb = $ouput_path.$fname.'_tn.jpg';
$cmd = "$ffmpeg -i $v_file -an -y -f mjpeg -ss 00:00:04 -vframes 1 $thmb";
$stat = system ($cmd);
}
Try with absolute paths in commands instead of depending on PATH ENV variable:
Both exec() and system() works. Resolve the path definitions.
/* Using Absolute paths */
$ffmpeg = "c:/wamp/www/includes/ffmpeg/bin/ffmpeg";
$videos = "c:/wamp/www/files/videos/*.mp4";
$ouput_path = "c:/wamp/www/files/thumbnails/";
foreach(glob($videos) as $video_file){
$filename = basename($video_file, ".mp4");
$thumbnail = $ouput_path.$filename.'_tn.jpg';
$command = "$ffmpeg -i $video_file -an -y -f mjpeg -ss 00:00:04 -vframes 1 $thumbnail";
$status = system ($command);
/*or
$status = exec($command);
if ($status === false) {
var_dump("ERROR: Conversion Failed!!!!");
} else
var_dump($status);
*/
}
I´m not allowed to add a comment, so here (as a Post) is what I found out while dealing with this:
The white space in the file name prevents ffmpeg from finding the right file. So one way is to change the file name.