How to hide "system()" output?

[opt_inf_env]

Hallo,

I want to see from php-file list of files which are in directory, where
is the given php-file. To do so, I put in the php-file the following
line:

<?php $sys = system("ls") ?>

Then I would like to work with string $sys. However, in the place where
the "system()" is executed, php-file print on screen result of the
execution. I would like to avoid this. Do you know how I can do it?

[NC]

[e-mail] wrote:

I want to see from php-file list of files which are in directory,
where is the given php-file. To do so, I put in the php-file the
following line:

<?php $sys = system("ls") ?>

Then I would like to work with string $sys. However, in the place
where the "system()" is executed, php-file print on screen result
of the execution. I would like to avoid this. Do you know how I
can do it?

$sys = `ls`;

Note that the command is enclosed in backticks (`), not single
quotes (').

Also, you may want to avoid program execution calls, because
they may not be available on your hosting service. Consider
using opendir()/readdir() instead:

http://www.php.net/opendir

Cheers,
NC

[Andy Hassall]

On 6 Apr 2005 13:36:15 -0700, op*********@yahoo.com wrote:

I want to see from php-file list of files which are in directory, where
is the given php-file. To do so, I put in the php-file the following
line:

<?php $sys = system("ls") ?>

Then I would like to work with string $sys. However, in the place where
the "system()" is executed, php-file print on screen result of the
execution. I would like to avoid this. Do you know how I can do it?

In this case, use opendir() and readdir() instead.

Or to answer the specific question, read the manual for system(), which points
to the related function exec(). But don't use that.

--
Andy Hassall / <an**@andyh.co.uk> / <http://www.andyh.co.uk>
<http://www.andyhsoftware.co.uk/space> Space: disk usage analysis tool