使用 grep 搜索包含点的字符串 [英] Using grep to search for a string that has a dot in it

问题描述

我正在尝试使用命令搜索字符串 0.49(带点)

I am trying to search for a string 0.49 (with dot) using the command

grep -r "0.49" *

但是发生的事情是我也得到了不需要的结果，其中包含诸如 0449、0949 等字符串.事情是 linux 将 dot(.) 视为任何字符并带出所有结果.但我只想得到0.49"的结果.

But what happening is that I am also getting unwanted results which contains the string such as 0449, 0949 etc,. The thing is linux considering dot(.) as any character and bringing out all the results. But I want to get the result only for "0.49".

推荐答案

grep 使用正则表达式；. 在正则表达式中表示任何字符".如果您需要文字字符串，请使用 grep -F、fgrep，或将 . 转义为 ..

grep uses regexes; . means "any character" in a regex. If you want a literal string, use grep -F, fgrep, or escape the . to ..

不要忘记用双引号将字符串括起来.否则你应该使用 \.

Don't forget to wrap your string in double quotes. Or else you should use \.

因此，您的命令需要是:

So, your command would need to be:

grep -r "0.49" *

或

grep -r 0\.49 *

或

grep -Fr 0.49 *

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