03-08-2010
it doesnt work with double quotes
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1. The problem statement, all variables and given/known data:
Hi,
I am trying to substitute an attributes value using sed command but it is only possible to substitute the value without space.
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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)