How to count all the lines of code in a directory recursively?

Question

We've got a PHP application and want to count all the lines of code under a specific directory and its subdirectories. We don't need to ignore comments, as we're just trying to get a rough idea.

wc -l *.php

That command works great within a given directory, but ignores subdirectories. I was thinking this might work, but it is returning 74, which is definitely not the case...

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

What's the correct syntax to feed in all the files?

Answer 1

You didn't specify how many files are there or what is the desired output. Is this what You are looking for:

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

?

Answer 2

Try

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

This may help as well

http://www.dwheeler.com/sloccount/

It'll give an accurate source lines of code count for whatever hierarchy you point it at, as well as some additional stats

Answer 3

what you want is a simple for loop:

total_count=0
for file in $(find . -name *.php -print)
do
count=$(wc -l $file)
let total_count+=count
done
echo $total_count

Answer 4

There is a little tool called sloccount to count the lines of code in directory. It should be noted that it does more than you want as it ignores empty lines/comments, groups the results per programming language and calculates some statistics.

Answer 5

For another one-liner:

( find ./ -name '*.php' -print0 | xargs -0 cat ) | wc -l

works on names with spaces, only outputs one number

Answer 6

very simply

find /path -type f -name "*.php" | while read FILE
do
    count=$(wc -l < $FILE)
    echo "$FILE has $count lines"
done

Answer 7

cat `find . -name "*.php"` | wc -l