Sed regexp 寻找空白或行尾 [英] Sed regexp looking for either whitespace or end of line

问题描述

我正在尝试检测包含三个部分的模式:

I'm trying to detect a pattern that has three parts:

1. 一个空格
2. m"或t"
3. 空格或行尾

我想保留#2 和#3.例如，我想将 我确定他没有" 更改为 我确定他没有"

I want to keep #2 and #3. For example, I'd like to change "i m sure he doesn t" to "im sure he doesnt"

我在表达 #3 时遇到问题，因为 [ $] 似乎只匹配空格，而不是行尾.这是我尝试过的:

I'm having trouble expressing #3, since [ $] only seems to match spaces, not line-ends. Here's what I've tried:

$ echo "i m sure he doesn t" | sed 's/ \([mt]\)\([ $]\)/\1\2/g'
im sure he doesn t

我应该如何在上面的表达式中表达空格或行尾"?谢谢！

How should I express "either a space or end of line" in the expression above? Thanks!

推荐答案

只匹配空格，然后是 m 或 t，然后空格或换行符不会捕获带有标点符号的情况，例如"please don t!" 中缺少 '.一个更通用的解决方案是改用词边界:

Just matching space, then m or t, then space or newline won't catch cases with punctuation, e.g. a missing ' in "please don t!". A more general solution is to use word boundaries instead:

echo "i m sure he doesn t test test don t." | sed 's/ \([mt]\)[[:>:]]/\1/g'

时髦的 [[:>:]] 在 OS X(我使用)上是必需的，请参阅 Larry Gerndt 对 sed 全词搜索和替换.在其他 sed 版本上，您可以使用 \b(任何单词边界)或 \> 代替.

The funky [[:>:]] is required on OS X (which I use), see Larry Gerndt's answer to sed whole word search and replace. On other sed flavors you may be able to use \b (any word boundary) or \> instead.

# example with word boundary
echo "i m sure he doesn t test test don t." | sed 's/ \([mt]\)[[:>:]]/\1/g'
im sure he doesnt test test dont.

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