Shell Programming and Scripting

Hi,

My script runs fine from the command line, but when crontab tries to launch it encounters lack of permissions:

could not open /dev/kbd to get keyboard type US keyboard assumed
could not get keyboard type US keyboard assumed
split: Permission denied

This is the head of the script:

Code:

#!/bin/ksh

if [ $# -ne 1 ]
then
echo "File needed as a parameter"
exit 0
fi
dos2unix $1 $1


hour=`date "+%H:%M:%S"`

/usr/bin/split -l 2000 $1
archi=`ls xa*|xargs`
...

Rest of the script skipped, it fails when trying to run a simple "split". Any ideas?

Sorry for the misleading title, I can't seem to edit it

There is no terminal when you run a cron job. You get the output - normally - in an email.

$1 if it is a file name, should be a fully qualified file name: /path/to/my/filename.
Either that or add a line before split:

Code:

cd /path/to/my

Completely agree. The same thing happened to me not too long ago. The explanation is simple: cron "needs to know" where to look to find the argument for the script. And it needs an absolute pathname (beginning with / down to its specific location in the filesystem tree).
On the other hand, when you run the script manually, you can either specify an absolute or relative pathname (a relative pathname starts with your current working directory), and that's why it's working when you run it from the command line and not from cron.
Adding another 2 cents here:
If the number of arguments is not equal to 1, the script should exit with error status (exit 1 instead of exit 0, which is used to indicate success).
If you're getting a "Permission denied" message, try adding your user name to your crontab (crontab -e):

Code:

USER=gacanepa

right before the cron jobs entry section.
Let us know how it goes.