What is the different between cron executing and shell execution?

[h.wulff <zuhause@aol.com>]

Hello,

I have a bash script which runs perfectly in the shell. But when cron 
executes the same script it causes errors.

if ( ! $SAV_INSTALL_PRG $SAV_INSTALL_OPTIONS 1>/dev/null )
then
    echo "Unable to install the new engine"
    echo "Command: \"$SAV_INSTALL_PRG $SAV_INSTALL_OPTIONS\""
    LOG2SYSLOG "ERROR: Unable to install the new engine."
else
    cd $TMP
    /bin/rm -rf $TMP/$TMP_SAV_INSTALL
    /bin/rm -f $IDE_DIR/*.ide
    LOG2SYSLOG "Successfully installed new sophos antivirus engine"
fi

All variables are defined in the header of this script.
and LOG2SYSLOG looks like this:

LOG2SYSLOG()
{
    MY_NAME=`basename $0`
    MY_PID=`pidof -x $MY_NAME`

    logger -t "$MY_NAME[$MY_PID]" $1
}

The error message is this:

Unable to install the new engine
Command: "/tmp/sav-install/install.sh -ni"
logger: invalid option -- n
usage: logger [-is] [-f file] [-p pri] [-t tag] [-u socket] [ message 
... ] 

I never call logger with a "n" option rather I call $SAV_INSTALL_PRG 
with an "n" option.
I dont see why logger get the "n" option...

Does anyone see the error?
Thank you for your efforts.
-- 

	h.wulff
	   [dont send me an email]