08-04-2014
Some more comments on top of what Chubler_XL said:
Just from looking at the snippet one can see many syntax errors, maybe sloppy typing. Make sure your variables are correctly expanded with the "$" sign in front. Surround "[", "]", and "=" with spaces. When quoting variables, make sure the "$" is inside the quotes, and don't use single quotes when you want expansion. "if If" will not work. And, some proper indentation will make the code more readable and safer.
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LEARN ABOUT PHP
escapeshellarg
ESCAPESHELLARG(3) 1 ESCAPESHELLARG(3)
escapeshellarg - Escape a string to be used as a shell argument
SYNOPSIS
string escapeshellarg (string $arg)
DESCRIPTION
escapeshellarg(3) adds single quotes around a string and quotes/escapes any existing single quotes allowing you to pass a string directly
to a shell function and having it be treated as a single safe argument. This function should be used to escape individual arguments to
shell functions coming from user input. The shell functions include exec(3), system(3) and the backtick operator.
On Windows, escapeshellarg(3) instead removes percent signs, replaces double quotes with spaces and adds double quotes around the string.
PARAMETERS
o $arg
- The argument that will be escaped.
RETURN VALUES
The escaped string.
EXAMPLES
Example #1
escapeshellarg(3) example
<?php
system('ls '.escapeshellarg($dir));
?>
SEE ALSO
escapeshellcmd(3), exec(3), popen(3), system(3), backtick operator.
PHP Documentation Group ESCAPESHELLARG(3)