Executing SASS script from PHP and getting output

Question

I have this PHP script (test.php):

<?php

    $cmd = "/usr/bin/sass --watch file1.scss";

    system($cmd);

?>

Now I call my PHP script from CLI this way:

/usr/bin/php test.php

And I get no output (it should print SASS is watching for changes).

If I call the SASS command directly from the shell, it outputs correctly.

Why?

Info: I'm using the PHP 5.3.6 version on OS X Lion

Edit: Please, note that this command watches for changes, it seems to behave differently to a regular command.

Edit2: The command works, it compiles correctly. The only thing lacking is the output (I want to debug and see errors :)

David Morales · Accepted Answer · 2012-01-18 10:45:24Z

1

Some command line utilities like sass, manipulate the output pipe in some way that PHP can't use.

So, in this particular case, there is no solution.

answered Jan 18, 2012 at 10:45

David Morales

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Massimiliano Arione · Accepted Answer · 2011-12-22 13:37:59Z

0

system() returns a string. Just echo it.

answered Dec 22, 2011 at 13:37

Massimiliano Arione

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2 Comments

Massimiliano Arione Over a year ago

so, it could be a problem with your PATH. Try do use the full path of sass executable (get it with "which sass")

David Morales Over a year ago

still not working, I'll update the question making clear that the command works, but it only lacks the output

powerbuoy · Accepted Answer · 2011-12-22 13:52:03Z

0

According to http://se.php.net/system you need to pass in a second argument to system() and the return value of the command will be set in that variable:

<?php
system($command, $return);

echo $return;

answered Dec 22, 2011 at 13:52

powerbuoy

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1 Comment

David Morales Over a year ago

this does not work either, please note that it is a command that executes itself in a loop watching for changes, so, I'm not searching for a return value, just the normal output (by the way, finalizing the command with ctrl+c outputs nothing in this case)

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