08-31-2009
Got it! Thanks again, Dr. House. When I was trying to figure this out, I would always include one in quotes, but not the other, so I could never get it to work. I did have to tweak it a bit, because my rookie scripting does things that probably should not be done... I had one variable inside of another.
So, I had:
Variable1=$1
Variable2=xxxx.$1.xxxx
In addition to putting the argument in Cron in quotes, and adding the $1 in quotes, I also had to go through the script, and every occurrence of Variable2 had to be put in quotes. I couldn't see it by looking at it, and being new enough that I didn't understand BASH variable expansion and such, but I kind of just stumbled on it by looking at the files that the script was creating, and wondering why I had two log files ('mediaserver.Bill' and 'Info.remotebackup.errors.txt') instead of one long-filenamed error log.
Thanks for the push in the right direction!
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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)