Grep --exclude-dir (root only)

Question

I am trying to configure a grep command that searches for my current directory but excludes the directory only if it is the root directory.

So, for the following directories, I want # 1 to be excluded and # 2 to be included

1) vendor/phpunit 2) app/views/vendor 

First I started with the command below

 grep -Ir --exclude-dir=vendor keywords * 

I tried using ^ vendor, ^ vendor /, ^ vendor /, ^ vendor, but nothing works.

Is there any way to do this with grep? I tried to try to do this with a single grep call, but if I need to, I can pass the results to a second grep.

Answer 1

With pipes:

 grep -Ir keywords * | grep -v '^vendor/' 

The problem with exclude-dir is that it checks the directory name and not the path to its entry, so it is not possible to distinguish between two vendor directories based on their depths.

Answer 2

Here is the best solution that will actually ignore the specified directory:

 function grepex(){ excludedir="$1" shift; for i in *; do if [ "$i" != "$excludedir" ]; then grep $@ "$i" fi done } 

You use it as a replacement in grep, just select the excluded directory as the first argument and leave * at the end. So your command will look like this:

 grepex vendor -Ir keywords 

This is not ideal, but as long as you don't have really weird folders (e.g. with type names -- or something else), it will cover most use cases. Feel free to refine it if you want something more complex.