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Top Forums Shell Programming and Scripting Getting error in command line arguments Post 302181478 by ag79 on Thursday 3rd of April 2008 04:58:57 AM
Old 04-03-2008
this is wrong
if [[ $1 != "dm5admin" ]]
here's what is going wrong: shell cant understand the following when u dont pass anything:
if [[ != "dm5admin" ]]


this is what you need
if [ "$1" != "dm5admin" ]

the shell will see this when u dont pass anything
if [ "" != "dm5admin" ]

which it will understand.

*Always* use double quotes when using shell script variable, unless you have a *very* good reason not to.

i dont know what you're trying to do with this script, but putting a password in it is not a great idea. anyway, thats your business.
 

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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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