count (non-blank) lines-of-code in bash

Question

In Bash, how do I count the number of non-blank lines of code in a project?

Answer 1

awk '/^[[:space:]]*$/ {++x} END {print x}' "$testfile"

Answer 2

cat file.txt | awk 'NF' | wc -l

Answer 3

cat foo.c | sed '/^\s*$/d' | wc -l

And if you consider comments blank lines:

cat foo.pl | sed '/^\s*#/d;/^\s*$/d' | wc -l

Although, that's language dependent.

Answer 4

grep -cvE '(^\s*[/*])|(^\s*$)' foo

-c = count
-v = exclude
-E = extended regex
'(comment lines) OR (empty lines)'
where
^    = beginning of the line
\s   = whitespace
*    = any number of previous characters or none
[/*] = either / or *
|    = OR
$    = end of the line

I post this becaus other options gave wrong answers for me. This worked with my java source, where comment lines start with / or * (i use * on every line in multi-line comment).

Answer 5

Script to recursively count all non-blank lines with a certain file extension in the current directory:

#!/usr/bin/env bash
(
echo 0;
for ext in "$@"; do
    for i in $(find . -name "*$ext"); do
        sed '/^\s*$/d' $i | wc -l ## skip blank lines
        #cat $i | wc -l; ## count all lines
        echo +;
    done
done
echo p q;
) | dc;

Sample usage:

./countlines.sh .py .java .html

Answer 6

It's kinda going to depend on the number of files you have in the project. In theory you could use

grep -c '.' <list of files>

Where you can fill the list of files by using the find utility.

grep -c '.' `find -type f`

Would give you a line count per file.