How to call a bash script from php passing parameters

Question

I have a php script that execute a bash script. I try to pass parameters like this:

$script="/opt/lampp/htdocs/adapt.sh"
$file="/opt/lampp/htdocs/videos/video1.mp4"
$prefix="Test"

exec ('.$script.' '.$file.' '.$prefix.');

What's wrong? How can I pass the parameters?

You can already spot what is wrong by just looking on how your syntax is highlighted... — Felix Kling
– Felix Kling, Commented May 13, 2010 at 10:40

Jan Gorman · Accepted Answer · 2010-05-13 10:37:56Z

3

You have your dots in the wrong place, should read:

exec ( $script . ' ' . $file . ' ' . $prefix );

or more readable

exec( "$script $file $prefix" );

answered May 13, 2010 at 10:37

Jan Gorman

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Comments

acm · Accepted Answer · 2010-05-13 10:38:40Z

1

this is wrong:

exec ('.$script.' '.$file.' '.$prefix.');

be carefull with quotes :-)

exec ($script.' '.$file.' '.$prefix);

answered May 13, 2010 at 10:38

acm

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Pekka · Accepted Answer · 2010-05-13 10:43:35Z

1

I dont really understand what your question is, but your exec() call should look like this:

exec ($script.' '.$file.' '.$prefix);

If you accept parameters from outside (e.g. from a GET or POST parameter), be sure to use escapeshellarg() on the arguments for security reasons.

answered May 13, 2010 at 10:38

Pekka

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