How can I exclude directories from grep -R?

Question

I want to traverse all subdirectories, except the \"node_modules\" directory.

Answer 1

Frequently use this:

grep can be used in conjunction with -r (recursive), i (ignore case) and -o (prints only matching part of lines). To exclude files use --exclude and to exclude directories use --exclude-dir.

Putting it together you end up with something like:

grep -rio --exclude={filenames comma separated} \
--exclude-dir={directory names comma separated}  

Describing it makes it sound far more complicated than it actually is. Easier to illustrate with a simple example.

Example:

Suppose I am searching for current project for all places where I explicitly set the string value debugger during a debugging session, and now wish to review / remove.

I write a script called findDebugger.sh and use grep to find all occurrences. However:

For file exclusions - I wish to ensure that .eslintrc is ignored (this actually has a linting rule about debugger so should be excluded). Likewise, I don't want my own script to be referenced in any results.

For directory exclusions - I wish to exclude node_modules as it contains lots of libraries that do reference debugger and I am not interested in those results. Also I just wish to omit .idea and .git hidden directories because I don't care about those search locations either, and wish to keep the search performant.

So here is the result - I create a script called findDebugger.sh with:

#!/usr/bin/env bash
grep -rio --exclude={.eslintrc,findDebugger.sh} \
--exclude-dir={node_modules,.idea,.git} debugger .