bash - 等待脚本中的 bash 后台作业完成

Question

为了最大化 CPU 使用率(我在 EC2 的 Debian Lenny 上运行东西)我有一个简单的脚本来并行启 Action 业:

#!/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更多代码。

有办法吗？

Answer 1

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.