使用grep搜索具有点的字符串

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

我正在使用这个命令搜索字符串0.49(带点)

grep -r "0.49" *

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".

但是发生的是我也得到了不需要的结果它包含了字符串，比如0449 0949等等。问题是linux将dot(.)作为任何字符并显示所有结果。但我想要得到的结果只有“0。49”

8 个解决方案

#1

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

grep使用regex;。意思是regex中的“任何字符”。如果您想要一个文字字符串，请使用grep -F、fgrep或转义。\。

#2

grep -F -r '0.49' * treats 0.49 as a "fixed" string instead of a regular expression. This makes . lose its special meaning.

grep -F -r '0.49' *将0.49视为“固定”字符串，而不是正则表达式。这使得。失去其特殊意义。

#3

You need to escape the . as "0\.49".

你需要逃离。为“0 \报”。

A . is a regex meta-character to match any character(except newline). To match a literal period, you need to escape it.

一个。是一个regex元字符，用于匹配任何字符(除了换行)。要匹配一个文字周期，您需要转义它。

#4

Just escape the .

只是逃避。

grep -r "0\.49"

grep - r“0 \报”

#5

You can escape the dot and other special characters using \

您可以使用\来转义点和其他特殊字符

eg. grep -r "0\.49"

如。grep - r“0 \报”

#6

Escape dot. Sample command will be.

逃逸点。示例命令。

grep '0\.00'

#7

You can also use "[.]"

您还可以使用“[.]”

grep -r "0[.]49"

#8

There are so many answers here suggesting to escape the dot with \. but I have been running into this issue over and over again: \. gives me the same result as .

这里有太多的答案建议用\来逃避点。但我已经反复地遇到这个问题:\。给我同样的结果。

However, these two expressions work for me:

然而，这两个表达对我来说是适用的:

$ grep -r 0\\.49 *

And:

和:

$ grep -r 0[.]49 *

I'm using a "normal" bash shell on Ubuntu and Archlinux.

我在Ubuntu和Archlinux上使用“普通”bash shell。