bash脚本-检查多个工作进程

Question

进程名称遵循特定的约定worker1.gs，其中X是一个序列号，值从1到4。我在那里有一个检查worker 1和worker 2的代码片段，依此类推……另外，下面的脚本在检查第一个process..how后终止，以修复它吗？

  #!/bin/bash

  pidof /usr/bin/gsx /var/app/transcoding/worker1.gs >/dev/null
  if [[ $? -ne 0 ]] ; then
    echo "Restarting Worker1: $(date)" >> /var/log/httpd/watchdog.txt
    /usr/bin/gsx /var/app/transcoding/worker1.gs
  fi

  pidof /usr/bin/gsx /var/app/transcoding/worker2.gs >/dev/null
  if [[ $? -ne 0 ]] ; then
    echo "Restarting Worker2: $(date)" >> /var/log/httpd/watchdog.txt
    /usr/bin/gsx /var/app/transcoding/worker2.gs
  fi

Answer 1

您可以像在C中一样创建循环：

for((i=0;i<10;i++)); do echo "$i" ; done

你可以简化

pidof /usr/bin/gsx /var/app/transcoding/worker1.gs >/dev/null
if [[ $? -ne 0 ]] ; then ... fi

转到

pidof /usr/bin/gsx /var/app/transcoding/worker1.gs >/dev/null || { ... }

合并：

for((i=0;i<10;i++)); do
  pidof /usr/bin/gsx /var/app/transcoding/worker1.gs >/dev/null || {
     echo "Restarting Worker$i: $(date)" >> /var/log/httpd/watchdog.txt
     /usr/bin/gsx /var/app/transcoding/worker$i.gs
  }   
done