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

Hi. As rich as the "find" command is, I fon't think there's an option

that asks "find" to ignore a specific directory.

But, we're talk "shell" here - there's always a way out! ;-)

1) you can always pipe the output of find into a filter - such as this:

find . -name "*.js" | grep -v "^UnwantedDir"

2) Since you're already using 'find' as part of a loop, you can always

do something programmatic like this:

for file in $(find . -name "*.js" )

do

[[ "$file" =~ ^./UnwantedDir.* ]] && continue

## get your work done

done

3) create a script to do the filtering and the actual work to be completed, and call that

script from the find "-exec" option ... as follows:

find . -name "*.js" -exec doSomeWork {} \;

So you can compensate for the fact that "find" may not have the option

you're looking for with a bit of ingenuity ...

Answer 2

the find command also has an "-exec" option. (execute). You might use that option to include a "grep".

Or why not just pipe the output to grep?

Example:

find . -name '*.js' | grep -v "<the file you DONT want to see>"

(the "-v" is the negative of)

so if you will want to do the loop,

for file in $(find . -name "*.js" | grep -v "blah blah");

do

java -jar config/yuicompressor-2.4.2.jar --type js $file -o $file;

done;