如何用grep搜索整个单词？ [英] How to search for whole words with grep?

问题描述

如果我有文件 foo ：

read_from_buffer
read_from_buffer_and_file
write_to_buffer
some_other_function

然后使用

cat foo | grep 'read_from_buffer'

会列出2行：

will list 2 lines:

read_from_buffer
read_from_buffer_and_file

但我想只有完全匹配...如何判断 grep ：不同的字符必须来自字符： 0-9a-zA-Z _

But I want only exact matches... How to tell grep that different character must come than character: 0-9a-zA-Z_

推荐答案

使用这个：

Use this:

grep -w 'read_from_buffer' foo

---

从 man grep ：

-w， - word-regexp ：只选择那些包含全部词的匹配项的行。测试是，匹配的子字符串必须是
或者位于行的开头，或者是以非单词组成字符开头。同样，它必须是
，或者在行尾，或者后面跟着一个非单词组成字符。单词组成字符是
字母，数字和下划线。

-w, --word-regexp: Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character. Similarly, it must be either at the end of the line or followed by a non-word constituent character. Word-constituent characters are letters, digits, and the underscore.

或

or

-x，--line-regexp ：仅选择与整行。 （-x由POSIX指定。）

-x, --line-regexp: Select only those matches that exactly match the whole line. (-x is specified by POSIX.)

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