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Old 08-18-2005, 03:56 AM   #1
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bash: run a script at most 3 times simultanously


I would like a script (let us call it simu) to run at most 3 times simultanously (to avoid overloading my server).

My idea would be to write a second script which would run simu only if less then 3 simus are already running. If 3 simus are running, he should wait for another one to finish, before starting a new one.

Does anyone know how I could do this?

Thank you for your help

Old 08-18-2005, 05:53 AM   #2
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Why not use a file to store a number and increment it following the number
of time you start your script ?
Old 08-18-2005, 06:38 AM   #3
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with xinetd you can sepcify how many servers to start.
dunno about inetd.
Old 08-18-2005, 10:43 AM   #4
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The problem you'll have is synchronisation. The solution has to be via some atomic operation, like a mv or a ln for example...

Old 08-18-2005, 03:49 PM   #5
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At the top of your script you could put:
check_instances() {
NUM_INSTANCES=$(ps aux | grep simu | wc -l)
if [[ $NUM_INSTANCES -ge 4 ]]; then
        echo "3 simu instances already running..."
        sleep 5

# the rest of your code here...
Here we check if your script is found 4 times in the process table rather than three, because "ps aux | grep simu | wc -l" is counted itself there. As written this function will check every 5 seconds, but you can change that by modifying the "sleep" value. Once there are less than 3 instances running the rest of your script will run...

There is a possibility for false-positives here (if the string "simu" is found elswhere in your process table) but this will dissappear if you choose a more unique name for your script.

Last edited by bulliver; 08-18-2005 at 03:53 PM.


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