How to get my expected search result using grep on Linux

Question

I am trying to find out which files contains this text ' roads ' on the server, so I use this command: grep -rl 'roads' . But it shows lots of such files:

./res/js/.svn/entries
./res/js/.svn/all-wcprops
./res/styles/.svn/entries
./res/styles/.svn/all-wcprops
./res/images/.svn/entries

I do not want this folder .svn show in the search result because this is just for version control, it means nothing for me, so is there a way that I can do this: if the result contains .svn , then it does not show up in the final result. eg below three files contain the text ' roads ':

 check.php 
./res/js/.svn/entries
./res/js/.svn/all-wcprops

Then the result only shows:

check.php

Answer 1

One simple way is to simply grep away your false positives:

grep -rl roads . | grep -v '/\.svn/'

If you want to be more efficient and not spend time searching through the SVN files, you can filter them away before grepping through them:

find -type f | grep -v '/\.svn/' | xargs grep -l roads

Answer 2

grep具有排除特定目录的功能，称为“--exclude-dir”，因此您只需将.svn作为“--exclude-dir”param传递，如下所示：

        "grep -rl --exclude-dir=.svn ./ -e roads"