如何使用SED更换[]支架 [英] How do i replace [] brackets using SED

问题描述

我有一个要删除标点符号的字符串.

I have a string that i am want to remove punctuation from.

我从

sed 's/[[:punct:]]/ /g'

但是我在HP-UX上遇到的问题不是一直都这样，有时在字符串中的$之后我会得到0和任何东西，但该内容会消失.所以我决定尝试手动进行.

But i had problems on HP-UX not liking that all the time, and some times i would get a 0 and anything after a $ in my string would dissappear. So i decided to try to do it manually.

我有以下代码可用于我感兴趣的所有标点符号，除了我似乎无法用其他任何东西在我的sed中添加方括号"[]"，否则它不会替代任何东西，而且我不会得到错误，所以我不确定要解决什么.

I have the following code which works on all my punctuation that I am interested in, except I cannot seem to add square brackets "[]" to my sed with anything else, otherwise it does not replace anything, and i dont get an error, so I am not sure what to fix.

无论如何，这是我目前拥有的，并且想要添加[].

Anyways this is what i currently have and would like to add [] to.

sed 's/[-=+|~!@#\$%^&*(){}:;'\'''\"''\`''\.''\/''\\']/ /g'

顺便说一句，我在Solaris，Redhat和amp;上使用KSH.惠普

BTW I am using KSH on Solaris, Redhat & HP

推荐答案

这是我最后得到的最终代码

Here is the final code I ended up with

`echo "$string" | sed 's/[^a-zA-Z0-9]/ /g'`

我必须将=和-放在最后.

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