GNU sed中的前瞻性正则表达式有什么问题? [英] What's wrong with my lookahead regex in GNU sed?

问题描述

这就是我正在做的事(简化示例):

This is what I'm doing (simplified example):

gsed -i -E 's/^(?!foo)(.*)$/bar\1/' file.txt

我试图将bar放在不是以foo开头的每一行的前面.这是错误:

I'm trying to put bar in front of every line that doesn't start with foo. This is the error:

gsed: -e expression #1, char 22: Invalid preceding regular expression

怎么了?

推荐答案

据我所知，sed既没有前瞻性也没有前瞻性.切换到具有类似语法的更强大的语言，例如perl.

As far as I know sed has not neither look-ahead nor look-behind. Switch to a more powerful language with similar syntax, like perl.

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