passing argument in cron job

I am new to cron, and need some advice. I currently have a shell script which runs like the example shown below. There are 4 of these duplicates I want to run. But I want to pass an arguement so that it changes to a different user-id within the script

0,5,10,15,20,25,30,35,40,45,50,55 ***** /../../../test.sh 2 [want to add an argument here for example $userid=xyz] >&1.

Now in this shell script there is a variable which needs to be changed when the cron job runs.

Example in shells script:

$user-id = (this needs to be defined from the cron job)


Can anyone help with this.

yes you need advice.

this is wrong:
0,5,10,15,20,25,30,35,40,45,50,55 *****
see man pages crontab
this is not really intelligent
0,5,10,15,20,25,30,35,40,45,50,55
and this is inpossible
/../../../test.sh

First never use relative paths in cron jobs i.e. /../../../test.sh
Always provide the full path i.e. /home/fpm/test.sh

Second, you can pass arguments to test.sh but it is up to
test.sh to interpret them as command line arguments.

Form a typical crontab(5) manpage ....

The ``sixth'' field (the rest of the line) specifies the
command to be run. The entire command portion of the
line, up to a newline or % character, will be executed by
/bin/sh or by the shell specified in the SHELL variable of
the cronfile. Percent-signs (%) in the command, unless
escaped with backslash (\), will be changed into newline
characters, and all data after the first % will be sent to
the command as standard input.

fpmurphy:
First never use relative paths in cron jobs i.e. /../../../test.sh
is that a relative path ?
IMO it's very absolute
pointing to an absolutely impossible path