How can I grep a keyword which contain underscore?

Question

I am currently searching some keyword something like

find -type f | xargs -grep -i -w 'weblogic_*'

But it show all the keyword match with weblogic instead of weblogic_

Answer 1

grep uses regular expressions, not globs (wildcard expressions).

In regular expressions, * is a quantifier that relates to the previous character or expression. Thus, _* is saying: zero or more instances of _, so NO _ will be matched as well.

You probably want:

'weblogic_.*'

which states that any (.) character may follow the _ zero or more times.

Note, however, that ending your regex in _.* partially contradicts grep's -w flag in that it will now only match the beginning of your regex on a word boundary.

If you wanted to be more explicit about this, you could use the word-boundary assertion \b and drop the -w option:

'\bweblogic_'

As you can see, this allows you to omit the .*, as grep performs substring matching by default, and you needn't match the remainder of the line if it is not of interest.

Also, there is no need for xargs: it is simpler and more efficient to use find's -exec primary, which has xargs built in, so to speak:

 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.

Finally, if your grep version supports the -R option, you can make do without find altogether and simply let grep process all files recursively:

 grep -R -i '\bweblogic_' .

Answer 2

When you use the pattern weblogic_*, it means look for weblogic followed by zero or more occurrences of _.

You can change it to use the pattern weblogic_.* if you want to avoid matching weblogic that is not followed by a _.

find -type f | xargs -grep -i -w 'weblogic_.*' 

should work.