Wait for bash background jobs in script to be finished


To maximize CPU usage (I run things on a Debian Lenny in EC2) I have a simple script to launch jobs in parallel:


for i in apache-200901*.log; do echo "Processing $i ..."; do_something_important; done &
for i in apache-200902*.log; do echo "Processing $i ..."; do_something_important; done &
for i in apache-200903*.log; do echo "Processing $i ..."; do_something_important; done &
for i in apache-200904*.log; do echo "Processing $i ..."; do_something_important; done &

I'm quite satisfied with this working solution, however I couldn't figure out how to write further code which only executed once all of the loops have been completed.

Is there a way to get control of this?

7/15/2009 1:44:15 PM

Accepted Answer

There's a bash builtin command for that.

wait [n ...]
      Wait for each specified process and return its termination  sta‐
      tus.   Each  n  may be a process ID or a job specification; if a
      job spec is given, all processes  in  that  job’s  pipeline  are
      waited  for.  If n is not given, all currently active child pro‐
      cesses are waited for, and the return  status  is  zero.   If  n
      specifies  a  non-existent  process or job, the return status is
      127.  Otherwise, the return status is the  exit  status  of  the
      last process or job waited for.
6/19/2017 9:39:04 AM

Using GNU Parallel will make your script even shorter and possibly more efficient:

parallel 'echo "Processing "{}" ..."; do_something_important {}' ::: apache-*.log

This will run one job per CPU core and continue to do that until all files are processed.

Your solution will basically split the jobs into groups before running. Here 32 jobs in 4 groups:

Simple scheduling

GNU Parallel instead spawns a new process when one finishes - keeping the CPUs active and thus saving time:

GNU Parallel scheduling

To learn more:

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