bash script to run a constant number of jobs in the background

Question

I need a bash script to run some jobs in the background, three jobs at a time.

I know can do this in the following way, and for illustration, I will assume the numbe

Answer 1

Maybe this could assist..

Sample usecase: run 'sleep 20' 30 times, just as an example. It could be any job or another script. Our control logic is to keep checking whether "how many already fired?" is less than or equal to "max processes defined", inside a while loop. If not, fire one and if yes, sleep .5 seconds.

Script output: In the below snip, it is observed that now we have 30 'sleep 20' commands running in the background, as we configured max=30.

%_Host@User> ps -ef|grep 'sleep 20'|grep -v grep|wc -l
30
%_Host@User>

Change value of no. of jobs at runtime: Script has a param "max", which takes value from a file "max.txt"(max=$(cat max.txt)) and then applies it in each iteration of the while loop. As seen below, we changed it to 45 and now we have 45 'sleep 20' commands running in the background. You can put the main script in background and just keep changing the max value inside "max.txt" to control.

%_Host@User> cat > max.txt
45
^C
%_Host@User> ps -ef|grep 'sleep 20'|grep -v grep|wc -l
45
%_Host@User>

Script:

#!/bin/bash
#---------------------------------------------------------------------#
proc='sleep 20' # Your process or script or anything..
max=$(cat max.txt)  # configure how many jobs do you want
curr=0
#---------------------------------------------------------------------#
while true
do
  curr=$(ps -ef|grep "$proc"|grep -v grep|wc -l); max=$(cat max.txt)
  while [[ $curr -lt $max ]]
        do
    ${proc} &        # Sending process to background.
    max=$(cat max.txt) # After sending one job, again calculate max and curr
    curr=$(ps -ef|grep "$proc"|grep -v grep|wc -l)
  done
  sleep .5    # sleep .5 seconds if reached max jobs.
done
#---------------------------------------------------------------------#

Let us know if it was any useful.