Linux 使用 grep 搜索包含点的字符串
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Using grep to search for a string that has a dot in it
提问by Varun kumar
I am trying to search for a string 0.49(with dot) using the command
我正在尝试0.49使用命令搜索字符串（带点）
grep -r "0.49" *
But what happening is that I am also getting unwanted results which contains the string such as 0449, 0949etc,. 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".
但是发生的事情是我也得到了不需要的结果，其中包含诸如0449,之类的字符串0949。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。
采纳答案by geekosaur
grepuses regexes; .means "any character" in a regex. If you want a literal string, use grep -F, fgrep, or escape the .to \..
grep使用正则表达式；.在正则表达式中表示“任何字符”。如果你想有一个文本字符串，使用grep -F，fgrep或逃避.到\.。
Don't forget to wrap your string in double quotes. Or else you should use \\.
不要忘记用双引号将字符串括起来。否则你应该使用\\.
So, your command would need to be:
所以，你的命令需要是：
grep -r "0\.49" *
or
或者
grep -r 0\.49 *
or
或者
grep -Fr 0.49 *
回答by Joni
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.它失去了它的特殊意义。
回答by Kevin Davis
You can escape the dot and other special characters using \
您可以使用 \ 转义点和其他特殊字符
eg. grep -r "0\.49"
例如。grep -r "0\.49"
回答by codaddict
You need to escape the .as "0\.49".
你需要逃避.as "0\.49"。
A .is a regex meta-character to match any character(except newline). To match a literal period, you need to escape it.
A.是一个正则表达式元字符，用于匹配任何字符（换行符除外）。要匹配文字句点，您需要对其进行转义。
回答by tuxuday
Escape dot. Sample command will be.
逃生点。示例命令将是。
grep '0\.00'
回答by cppcoder
Just escape the .
只是逃避 .
grep -r "0\.49"
grep -r "0\.49"
回答by skan
You can also use "[.]"
您也可以使用“[.]”
grep -r "0[.]49"
回答by Afr
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 .
这里有这么多的答案，提示逃生用点\.，但我一直在一遍又一遍地运行到这个问题：\.给我的结果相同.
However, these two expressions work for me:
但是，这两个表达式对我有用：
$ 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”的 bash shell。
回答by smc
You can also search with -- option which basically ignores all the special characters and it won't be interpreted by grep.
您还可以使用 -- 选项进行搜索，该选项基本上会忽略所有特殊字符，并且不会被 grep 解释。
$ cat foo |grep -- "0\.49"
