用空格替换行，并用包含空格的字符串替换反斜杠 [英] Replace line with space and backslash with a string containing spaces

问题描述

我要替换以下行:

--memory 20g \

使用

--memory 100g \

实际上，它应该替换--memory之后的任何数字.以下是我所拥有的，但无法获得预期的结果.

Actually it should replace any number after --memory. Following is what I have, but not able to get the expected result.

sed -i -E -- "s/\b--memory.*/--memroy 100g \/g"  a.txt

推荐答案

此处不需要扩展的正则表达式支持(-E)，只能通过POSIX进行以下操作.这个想法是，您需要对元字符\进行两次转义以使它成为文字

You don't need the extended regex support here (-E), POSIX-ly you could just do as below. The idea is you need to double-escape the meta-character \ to make it a literal

sed 's/--memory \(.*\) \\/--memory 100g \\/g' a.txt

，或者如果您确定它始终都是20g，请直接使用字符串.

or if you are sure its going to be 20g all the time, use the string directly.

sed 's/--memory 20g \\/--memory 100g \\/g' a.txt

使用\(.*\)的一个优点是，您可以替换该位置可能发生的任何事情. .*是一个贪婪的表达式，可以匹配任何内容，并且在POSIX sed(基本正则表达式)中，您需要将捕获的组转义为\(.*\)，而如果在启用-E标志的情况下执行相同的操作(在GNU/FreeBSD sed)，您可以执行(.*).如果要匹配精确的行，并且不让sed在不需要的地方替换文本，也可以使用正则表达式锚点^，$.与ERE相同的操作

The one advantage of using \(.*\) is that allows you to replace anything that could occur in that place. The .* is a greedy expression to match anything and in POSIX sed (Basic Regular Expressions) you need to escape the captured group as \(.*\) whereas if you do the same with the -E flag enabled (on GNU/FreeBSD sed) you could just do (.*). Also use regex anchors ^, $ if you want to match the exact line and not to let sed substitute text in places that you don't need. The same operation with ERE

sed -E 's/--memory (.*) \\/--memory 100g \\/g' file

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