Bash:与grep命令处于相反状态的单线退出吗? [英] Bash: One-liner to exit with the opposite status of a grep command?

问题描述

如何减少以下bash脚本?

How can I reduce the following bash script?

grep -P "STATUS: (?!Perfect)" recess.txt && exit 1
exit 0

似乎我应该可以用一个命令来完成，但是这里总共有3个.

It seems like I should be able to do it with a single command, but I have a total of 3 here.

我的程序应该:

• 阅读cess.txt
• 退出1(或非零)(如果其中包含一行"STATUS:"为NOT完美"的行)
• 如果不存在这样的行，则退出0(即所有"STATUS:"行均为"Perfect")

答案是最严格的脚本.谢谢！

The answer award goes to the tightest script. Thanks!

该文件的程序应具有退出状态0 :

FILE: styles.css 
STATUS: Perfect!

FILE: contour-styles.css
STATUS: Perfect!

此文件的程序应具有退出状态1 (或非零):

Program should have exit status 1 (or non-zero) for this file:

FILE: styles.css 
STATUS: Perfect!

FILE: contour-styles.css
STATUS: Busted 
FAILURES: 1 failure

Id's should not be styled
       1. #asdf

推荐答案

只需否定返回值.

! grep -P "STATUS: (?!Perfect)" recess.txt

这篇关于Bash:与grep命令处于相反状态的单线退出吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！