Shell Programming and Scripting

Hi All,

I have set a cronjob to run a script at a frequency of 10 mins.
However, if the script takes more than 10 mins to complete, how will the cronjob behaves?
1) Does it ignore the current process and restart the whole process again?
2) Or does it stop to trigger the script again unless previous process has been completed?

Depends on your cron implementation but generally it will run another copy of the script 10 mins after the first one is started - regardless of how long the first run takes.

Solutions:

Hi,

Running the script continously means to say that i have to leave a command prompt window always open. Once closed, the process will be ended, and also another disadvantage is that a infinite looping script will eat up alot of resources, wouldn;t it ?
So i still think cron is the more practical one. But i have no idea about anacron or control-m. Can you enlightened me on that ?

Can you give an example of script being re-entrant ?
I was thinking if i could make the script check for the process to see if the previous process have been completed, can any experts give me some examples of this ? I am using csh by the way.

Not at all, just move it to the background, no different to all the other daemonised stuff you are running on your system. If it likes to write to STDOUT, redirect this to a log file (or /dev/null if you prefer).
Much as before, no problems here either. Just use sleep at the bottom of the loop to stop it thrashing.
Your call of course.
Anacron is a slighty more advanced version of cron, may or may not do what you want. Free (I think)
Control-M is a BMC product that manages scheduling at an enterprise level, it supports job dependancies, understands limited resources and works across multiple platforms. Costs a bit.

Google for more info on these
Reentrant just means that it can be run multiple times at once without going wonky. One example:
Not rentrant:

Code:

#!/bin/sh
/usr/bin/do_some_stuff > /var/log/did_some_stuff.log
do some other things
if grep 'it worked' /var/log/did_some_stuff.log
then
  /usr/sbin/assume_we_are_good_to_go
else
  echo "Argh"
fi

Reentrant:

Code:

#!/bin/sh
/usr/bin/do_some_stuff > /var/log/did_some_stuff.$$.log
do some other things
if grep 'it worked' /var/log/did_some_stuff.$$.log
then
  /usr/sbin/assume_we_are_good_to_go
else
  echo "Argh"
fi
rm /var/log/did_some_stuff.$$.log

The first example could end up reading back the log of a different instance of itself. The second version includes the current process ID in the log file which prevents this happening.

Another, simpler, way is to just look in the process table for another copy of the script, if found exit immediatly.

(Assuming system V as opposed to bsd)

Code:

#!/bin/csh
set numprocs=`ps -ef | grep -v grep | grep $scriptname | wc -l | awk '{ print $1 }'`
if ($numprocs == 1) then
    # carry on
else
   echo "Another instance is already running, exiting..."
   exit 0
endif

Hi ,

I am refering to this particular exmaple that you have provided me.

Code:

#!/bin/csh
set numprocs=`ps -ef | grep -v grep | grep $scriptname | wc -l | awk '{ print $1 }'`
if ($numprocs == 1) then
    # carry on
else
   echo "Another instance is already running, exiting..."
   exit 0
endif

What is the significance of the numprocs?
If it equals to 1 , why does it show that the script is not running anymore?

I tried a simple csh script below and it doesn;t really work.
When it is sleeping, the name of the script which is "myscriptname" is not reflected during " ps -ef ", it only shows " usr 28951 28941 0 15:00:39 pts/7 0:00 sleep 100 ".

Can you help ?

Code:

#!/bin/csh

echo xxx
echo ppp 
echo zzz

sleep 100

Because one copy is running, the one that's currently executing. It's looking for another version of itself (ie 2 or more) in case there are other copies still running.
Works for me, when I run a csh script with a sleep in it, I see the sleep, _and_ the main script running.