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Top Forums UNIX for Dummies Questions & Answers shell script programming issues Post 302263164 by sivakumar.rj on Monday 1st of December 2008 01:58:37 AM
Old 12-01-2008
No need to have the quotes in "$#" -ne "1"
if [ ! -d $1 ] here to have the quotes...for $1...

Kindly let me know what is the error u r getting...so that it will be easier to solve...
 

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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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