PHP exec() not grabbing output from bash command

Question

i'm trying to get the ouput from a bash command. If I execute the some command directly in the bash, i see the result. But not with PHP exec() function..

$command = "ffmpeg -i '$video_path' 2>&1 | grep Video | perl -wle 'while(<>){ $_ =~ /.*?(\d+x\d+).*/; print $1; }'";
$res = exec($command/*, $output*/);

print_r($res);
// print_r($output);

This command try to get a video resolution using ffmpeg lib. The video path it's fine, I triple check that.

Thanks!

EDIT: screenshot from the output in bash

Per the docs, "The last line from the result of the command. If you need to execute a command and have all the data from the command passed directly back without any interference, use the passthru() function." — ceejayoz
– ceejayoz, Commented Aug 15, 2017 at 15:29
The easiest way to get the full output of a command into a string is to use backticks (`) — iainn
– iainn, Commented Aug 15, 2017 at 15:30
It looks like you have tried to use the output argument to exec, which will contain all of the output, rather than the last line. What is in $output when you run it that way? — Rob Ruchte
– Rob Ruchte, Commented Aug 15, 2017 at 15:34
The output has just one line, I add an screenshot in the question where you can check that. — mauriblint
– mauriblint, Commented Aug 15, 2017 at 15:34
Should I need to escape some of this chars? $_ =~ /.*?(\d+x\d+).*/; ? — mauriblint
– mauriblint, Commented Aug 15, 2017 at 15:35

Rob Ruchte · Accepted Answer · 2017-08-15 15:47:06Z

2

Try escaping the $ in your regex with a backslash. Since you're wrapping everything in double quotes, PHP is trying to insert the value of $_, which is not a thing. I got an undefined variable notice when I ran your code. It worked for me when I escaped the $.

answered Aug 15, 2017 at 15:47

Rob Ruchte

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