用包含空格的字符串替换空格和反斜杠行

I want to replace the following line:

我想替换以下行：

--memory 20g \

with

同

--memory 100g \

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

实际上它应该在--memory之后替换任何数字。以下是我所拥有的，但无法获得预期的结果。

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

1 个解决方案

#1

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

你不需要这里的扩展正则表达式支持（-E），POSIX-ly你可以这样做。这个想法是你需要双重转义元字符\才能使它成为文字

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

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

或者如果你确定它一直是20克，直接使用字符串。

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

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

使用\（。* \）的一个优点是允许您替换在该位置可能发生的任何事情。。*是一个贪婪的表达式来匹配任何东西，在POSIX sed（基本正则表达式）中你需要将捕获的组转义为\（。* \），而如果你在启用-E标志时也这样做（在GNU / FreeBSD上） sed）你可以做（。*）。如果你想匹配确切的行而不是让sed在你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

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

#1