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Full Discussion: Cron job is not running
Operating Systems Solaris Cron job is not running Post 302597374 by SunilB2011 on Friday 10th of February 2012 05:48:57 AM
Old 02-10-2012
Quote:
Originally Posted by rbatte1
Probably a silly question, but is the script executeable by the user defined with the cron job?

As an aside, it is an odd definition you have because you explicitly say every minute with the range 0-59. Would an * not suffice?



Anyway, just a thought.


Robin
Liverpool/Blackburn
UK

The script is executing fine when running it as a root user from the command line.

---------- Post updated at 04:18 PM ---------- Previous update was at 04:16 PM ----------

Quote:
Originally Posted by itkamaraj
redirect your stderr and stdout

Code:
0-59 * * * * /var/tmp/r.sh 1>/tmp/out.txt 2>&1

Got the following error.

Code:
 
/var/tmp/r.sh: syntax error at line 1: `dt1=$' unexpected

But the above error is not thrown when running the script from the command line.
 

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SHELL-QUOTE(1)						User Contributed Perl Documentation					    SHELL-QUOTE(1)

NAME
shell-quote - quote arguments for safe use, unmodified in a shell command SYNOPSIS
shell-quote [switch]... arg... DESCRIPTION
shell-quote lets you pass arbitrary strings through the shell so that they won't be changed by the shell. This lets you process commands or files with embedded white space or shell globbing characters safely. Here are a few examples. EXAMPLES
ssh preserving args When running a remote command with ssh, ssh doesn't preserve the separate arguments it receives. It just joins them with spaces and passes them to "$SHELL -c". This doesn't work as intended: ssh host touch 'hi there' # fails It creates 2 files, hi and there. Instead, do this: cmd=`shell-quote touch 'hi there'` ssh host "$cmd" This gives you just 1 file, hi there. process find output It's not ordinarily possible to process an arbitrary list of files output by find with a shell script. Anything you put in $IFS to split up the output could legitimately be in a file's name. Here's how you can do it using shell-quote: eval set -- `find -type f -print0 | xargs -0 shell-quote --` debug shell scripts shell-quote is better than echo for debugging shell scripts. debug() { [ -z "$debug" ] || shell-quote "debug:" "$@" } With echo you can't tell the difference between "debug 'foo bar'" and "debug foo bar", but with shell-quote you can. save a command for later shell-quote can be used to build up a shell command to run later. Say you want the user to be able to give you switches for a command you're going to run. If you don't want the switches to be re-evaluated by the shell (which is usually a good idea, else there are things the user can't pass through), you can do something like this: user_switches= while [ $# != 0 ] do case x$1 in x--pass-through) [ $# -gt 1 ] || die "need an argument for $1" user_switches="$user_switches "`shell-quote -- "$2"` shift;; # process other switches esac shift done # later eval "shell-quote some-command $user_switches my args" OPTIONS
--debug Turn debugging on. --help Show the usage message and die. --version Show the version number and exit. AVAILABILITY
The code is licensed under the GNU GPL. Check http://www.argon.org/~roderick/ or CPAN for updated versions. AUTHOR
Roderick Schertler <roderick@argon.org> perl v5.16.3 2010-06-11 SHELL-QUOTE(1)
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