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

fpmurphy, your post gave me the answer to my issue immediately. it's remarkable because you pasted in the part of the manpage that would help me (regarding %). if you hadn't done that, i'd still be looking.

i definitely need to start reading manpages first... they always surprise me with how informative they are.