How to exclude a directory in find . command

Question

I'm trying to run a find command for all JavaScript files, but how do I exclude a specific directory?

Here is the find code we're using.

for file in $(find . -name '*.js')
do 
  java -jar config/yuicompressor-2.4.2.jar --type js $file -o $file
done

Answer 1

This is the only one that worked for me.

find / -name NameOfFile ! -path '*/Directory/*'

Searching for "NameOfFile" excluding "Directory". Give emphasis to the stars * .

Answer 2

 find . -name '*.js' -\! -name 'glob-for-excluded-dir' -prune

Answer 3

For those of you on older versions of UNIX who cannot use -path or -not

Tested on SunOS 5.10 bash 3.2 and SunOS 5.11 bash 4.4

find . -type f -name "*" -o -type d -name "*excluded_directory*" -prune -type f

Answer 4

If -prune doesn't work for you, this will:

find -name "*.js" -not -path "./directory/*"

Answer 5

This is the format I used to exclude some paths:

$ find ./ -type f -name "pattern" ! -path "excluded path" ! -path "excluded path"

I used this to find all files not in ".*" paths:

$ find ./ -type f -name "*" ! -path "./.*" ! -path "./*/.*"

Answer 6

find -name '*.js' -not -path './node_modules/*' -not -path './vendor/*'

seems to work the same as

find -name '*.js' -not \( -path './node_modules/*' -o -path './vendor/*' \)

and is easier to remember IMO.