How to count all the lines of code in a directory recursively using or operator to include several file types

Question

I am reading this answer: How to count all the lines of code in a directory recursively?

But it only works for php files. I want to count recursively including javascript, html, css and php files to get the total number of combined lines of code.

I tried this find . -name '*.php' |'*.js' | xargs wc -l

But it does not work.

Note: I am using OS X El Capitan.

Answer 1

to avoid using xargs

find \( -name '*.php' -o -name '*.js' -o -name '*.html' \) -exec wc -l {} +

• \( \) grouping so that all results are given to wc command
• add as many -o -name as required within \( \)
• use iname instead of name to ignore case of filenames
• by default, find works on current directory

Reference:

• https://unix.stackexchange.com/questions/12902/how-to-run-find-exec
• https://unix.stackexchange.com/questions/24954/when-is-xargs-needed
• https://unix.stackexchange.com/questions/131766/why-does-my-shell-script-choke-on-whitespace-or-other-special-characters/131767#131767

Answer 2

Give this a go:

find . -name '*.php' | xargs wc -l

With include files:

find . -name '*.php' -o -name '*.js' | xargs wc -l

To not include certain path:

find . -name "*.php" -not -path "./tests*" | xargs wc -l

Answer 3

Here is a one liner. It uses find regex option to find for matching extensions, and wc --files0-from=- parameter to get file list from standard input

find -regex '.*\(php\|js\|css\|html?\)' -printf '%p\0' |wc -l --files0-from=-