Help, while loop and sed not producing desired output Post: 302483534

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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)

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Help me in getting the desired output

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