bash 如何 grep 包含下划线的关键字？
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How can I grep a keyword which contain underscore?
提问by taymedee
I am currently searching some keyword something like
我目前正在搜索一些关键字，例如
find -type f | xargs -grep -i -w 'weblogic_*'
find -type f | xargs -grep -i -w 'weblogic_*'
But it show all the keyword match with weblogicinstead of weblogic_
但它显示所有与weblogic而不是匹配的关键字weblogic_
回答by mklement0
grepuses regular expressions, not globs (wildcard expressions).
grep使用正则表达式，而不是 globs（通配符表达式）。
In regular expressions, *is a quantifier that relates to the previous character or expression. Thus, _*is saying: zero or moreinstances of _, so NO _will be matched as well.
在正则表达式中，*是与前一个字符或表达式相关的量词。因此，_*是说： 的零个或多个实例_，因此 NO_也将匹配。
You probably want:
你可能想要：
'weblogic_.*'
which states that any (.) character may follow the _zero or more times.
这表明任何 ( .) 字符都可以跟随_零次或多次。
Note, however, that ending your regex in _.*partially contradicts grep's -wflag in that it will now only match the beginningof your regex on a word boundary.
但是请注意，正则表达式的结尾_.*部分与grep的-w标志相矛盾，因为它现在只会匹配单词边界上正则表达式的开头。
If you wanted to be more explicit about this, you could use the word-boundary assertion \band drop the -woption:
如果你想更明确地说明这一点，你可以使用 word-boundary 断言\b并删除-w选项：
'\bweblogic_'
As you can see, this allows you to omit the .*, as grepperforms substringmatching by default, and you needn't match the remainder of the line if it is not of interest.
如您所见，这允许您省略.*, 因为默认情况下grep执行子字符串匹配，如果您不感兴趣，则不需要匹配该行的其余部分。
Also, there is no need for xargs: it is simpler and more efficient to use find's -execprimary, which has xargsbuilt in, so to speak:
此外，不需要xargs: 使用内置的find's -execprimary更简单、更高效xargs，可以这么说：
find . -type f -exec grep -i '\bweblogic_' {} +
{}represents the list of input filenames and +specifies that as many input filenames as possible should be passed at once - as with xargs.
{}表示输入文件名的列表，并+指定应一次传递尽可能多的输入文件名 - 与xargs.
Finally, if your grepversion supports the -Roption, you can make do without findaltogether and simply let grep process all files recursively:
最后，如果您的grep版本支持该-R选项，您可以完全不用find，只需让 grep 递归处理所有文件：
grep -R -i '\bweblogic_' .
回答by R Sahu
When you use the pattern weblogic_*, it means look for weblogicfollowed by zero or more occurrences of _.
当您使用该模式时weblogic_*，这意味着查找weblogic后跟零次或多次出现的_。
You can change it to use the pattern weblogic_.*if you want to avoid matching weblogicthat is not followed by a _.
weblogic_.*如果您想避免weblogic后面没有跟的匹配，您可以将其更改为使用模式_。
find -type f | xargs -grep -i -w 'weblogic_.*'
should work.
应该管用。
回答by Amit Verma
Try without regex
尝试不使用正则表达式
find . -type f | xargs grep -i 'weblogic_'
回答by Taher Khorshidi
simply use this :
简单地使用这个：
grep -i '^weblogic_.*'
