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使用 grep 搜索特定字符串

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使用 grep 搜索特定字符串

我想在文件中搜索特定的字符串并仅显示与该字符串完全匹配的行，但这似乎不起作用grep -w。

这只是一个例子，因为我将在脚本中搜索字符串，然后只想显示完全匹配的行，而不是“string-xxxx”或“string.io”。

有人有什么主意吗？

root@ubuntu-client:/var/lib/apt/lists# cat aa
aide is good
this is aide one
that is aide-common good
aide-good is not ok
hello aide-help
root@ubuntu-client:/var/lib/apt/lists# 

oot@ubuntu-client:/var/lib/apt/lists# grep -w aide aa
aide is good
this is aide one
that is aide-common good
aide-good is not ok
hello aide-help
root@ubuntu-client

答案1

您遇到的问题是被视为非单词字符，因此中的-字符串被视为单词。来自：aideaide-man grep

  -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.  This option has no effect if -x is also specified.

目前还不清楚你的实际要求是什么，但是如果你想匹配一个前面或后面没有任何非空白字符，你可以切换到 grep 的 PCRE 模式并执行以下操作

$ cat aa
aide is good
this is aide one
that is aide-common good
aide-good is not ok
hello aide-help

$ grep -P '(?<!\S)aide(?!\S)' aa
aide is good
this is aide one

但请注意，这必然也会排除其他尾随标点符号，如aide,和aide.。仅匹配aide前面或后面没有单词字符或连字符，你可以使用

grep -P '(?<!(\w|-))aide(?!(\w|-))' aa

如需参考，请参阅前瞻和后瞻零长度断言

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