等待脚本中的bash后台作业完成 [英] Wait for bash background jobs in script to be finished

问题描述

为了最大化CPU使用率(我在EC2中的Debian Lenny上运行东西)，我有一个简单的脚本来并行启动作业:

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

#!/bin/bash

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 &
...

我对这个可行的解决方案非常满意，但是我无法弄清楚如何编写仅在所有循环完成后才执行的其他代码.

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.

有没有办法控制这个?

推荐答案

有一个bash内置命令.

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.

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