输入不匹配时，防止grep返回错误 [英] Prevent grep returning an error when input doesn't match

问题描述

我想在bash脚本中写一段代码来检查程序是否已经运行。
我有以下内容来搜索bar是否正在运行

  foo =`ps -ef | grep bar | grep -v grep` 
  

  grep -v grep 
  

部分是为了确保grep在ps结果中不考虑bar

当bar未运行时，foo正确为空。但我的问题在于，脚本中有

  set -e 
  

如果某个命令返回错误，那么这是一个终止脚本的标志。
事实证明，当bar未运行时，grep -v grep与任何内容都不匹配，并且grep返回错误。我尝试使用-q或-s但无济于事。

有没有解决方法？ Thx

解决方案

当然：

  ps -ef | grep bar | {grep -v grep ||真正; } 
  

甚至：

  ps -ef | grep bar | grep -v grep | cat 
  

I want to write in a bash script a piece of code that checks if a program is already running. I have the following in order to search whether bar is running

 foo=`ps -ef | grep bar | grep -v grep`

The

 grep -v grep

part is to ensure that the "grep bar" is not taken into account in ps results

When bar isn't running, foo is correctly empty. But my problem lies in the fact tha the script has

 set -e

which is a flag to terminate the script if some command returns an error. It turns out that when bar isn't running, "grep -v grep" doesn't match with anything and grep returns an error. I tried using -q or -s but to no avail.

Is there any solution to that? Thx

解决方案

Sure:

ps -ef | grep bar | { grep -v grep || true; }

Or even:

ps -ef | grep bar | grep -v grep | cat

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