Shell Programming and Scripting

Hi,

I have an existing cron job like the following.

15 5 * * * /appl/scripts/MyScript.sh 2>/dev/null >/dev/null

and this works just fine.

Now I had to modify the script and now it requires an input parm as 2.

I had modified the cron entry so that the script can run now with the parm=2.

My new cron entry is below.

15 5 * * * /appl/scripts/MyScript.sh 2 2>/dev/null >/dev/null

Now the script is failing with a complaint that it could not find the input parm.

Any ideas ? on how to get around this problem.

Any help is appreciated.

Thanx..
-Bheem

The cron entry is specified correct. Therefore the problem seems to be within the script.

To tell what is going wrong there we would have to see the script.

When I run the script from command prompt, it runs fine.

MyScript.sh 2

Inside the script I am assigning $1 to a variable and using that to pass to a stored procedure.

I tried the following cron entries too, but none worked.

15 5 * * * /appl/scripts/MyScript.sh '2' 2>/dev/null >/dev/null

15 5 * * * /appl/scripts/MyScript.sh "2" 2>/dev/null >/dev/null

15 5 * * * /appl/scripts/MyScript.sh \2 2>/dev/null >/dev/null

Like I said, the mistake is inside the script.

But since you clearly don't want to show it, you can't be helped.

The parameter itself is not the problem.

Can you run the script as standalone successfully without cronjob. That way, atleast we know, it's not script problem

Hi roopla and sb008,

As I said earlier, there was no problem in the script. I could run it successfully as a standalone script.

Here is my complete script:
-------------------------------------------------
#!/usr/bin/ksh
. ~/.profile # for cron job environment visibility

filedate=`date +%m%d%H`
LogPath=/appl/loopqual/ait/log
Log=$LogPath/MyScript_$filedate.log
logtime=`date +%H%M%S`
echo "MyScript.sh started $logtime\n" > $Log

endmsg()
{
logtime=`date +%H%M%S`
echo "\nMyScript.sh end" $logtime >> $Log
exit 0
}

if [ $# -ne 1 ]
then
echo "Missing input parmameter\n" >> $Log
endmsg
fi
delete_agedmlr()
{
echo "Execute spDeleteAgedMlr" >> $Log
/appl/loopqual/ait/scripts/run_sp_2op.sh spDeleteAgedMlr W$1 delete_agedmlr

if [[ $? -ne 0 ]]
then
echo "spDeleteAgedMlr failed\n" >> $Log
fi
echo "spDeleteAgedMlr completed\n" >> $Log
}

delete_agedmlr $1
endmsg
sync
----------------------------------------------
Here is the output (from log file) when I ran it from cron.

MyScript.sh started 105331

Missing input parmameter


MyScript.sh end 105331

---------------------------------------------
Here is the output (from log file) when I ran it from command prompt.

$MyScript.sh 2

MyScript.sh started 105856

Execute spDeleteAgedMlr
spDeleteAgedMlr completed


MyScript.sh end 105857
------------------------------------------------

I do not understand what is going on here and why the script is not able to find the input parmater from cron entry.

Please help!!!!
Thanx..
-Bheem

I'm guessing that the command line in cron is reading all your redirection (to /dev/null) as parameters.

Either don't test for the number of passed parameters (get rid of the $# test), check that the number of passed parameters is greater than 0, or just check that $1 is a valid parameter.

Oh, and set the $1 variable early, not late, in case the $1 value is changed before you use it later on in your script.

That's my guess...

Actually, I'm not seeing where you set the variable to the contents of $1. Am I missing something?