从bash输出中排除一个字符串 [英] Exclude one string from bash output

问题描述

我正在研究一个项目.在这个项目中，由于某些原因，我需要从与模式匹配的输出(或文件)中排除第一个字符串.困难在于我只需要从流中排除一个字符串，仅排除第一个字符串.例如，如果我有:

I'm working now on a project. In this project for some reasons I need to exclude first string from the output (or file) that matches the pattern. The difficulty is in that I need to exclude just one string, just first string from the stream. For example, if I have:

1 abc
2 qwerty
3 open
4 abc
5 talk

一些脚本工作后，我应该有这个:

After some script working I should have this:

2 qwerty
3 open
4 abc
5 talk

注意:我对单词之前的数字一无所知，因此无法使用有关它们的知识来过滤输出.

NOTE: I don't know anything about digits before words, so I can't filter the output using knowledge about them.

我已经用grep编写了小脚本，但是它切出了与模式匹配的每个字符串:

I've written small script with grep, but it cuts out every string, that matches the pattern:

'some program' | grep -v "abc"

阅读有关awk，sed等的信息，但不知道我是否可以解决我的问题.有什么帮助，谢谢.

Read info about awk, sed, etc. but didn't understand if I can solve my problem. Anything helps, Thank you.

推荐答案

使用awk:

some program | awk '{ if (/abc/ && !seen) { seen = 1 } else print }'

或者，仅使用过滤器:

some program | awk '!/abc/ || seen { print } /abc/ && !seen { seen = 1 }'

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