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

Question

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.

#!/bin/bash
find . -path './pma' -prune -o -path './blog' -prune -o -path './punbb' -prune -o -path './js/3rdparty' -prune -o -print | egrep '\.php|\.as|\.sql|\.css|\.js' | grep -v '\.svn' | xargs cat | sed '/^\s*$/d' | wc -l

The above will give you the total count of lines of code (blank lines removed) for a project (current folder and all subfolders recursively).

In the above "./blog" "./punbb" "./js/3rdparty" and "./pma" are folders I blacklist as I didn't write the code in them. Also .php, .as, .sql, .css, .js are the extensions of the files being looked at. Any files with a different extension are ignored.

If you want to use something other than a shell script, try CLOC:

cloc counts blank lines, comment lines, and physical lines of source code in many programming languages. It is written entirely in Perl with no dependencies outside the standard distribution of Perl v5.6 and higher (code from some external modules is embedded within cloc) and so is quite portable.

There are many ways to do this, using common shell utilities.

My solution is:

grep -cve '^\s*$' <file>

This searches for lines in <file> the do not match (-v) lines that match the pattern (-e) '^\s*$', which is the beginning of a line, followed by 0 or more whitespace characters, followed by the end of a line (ie. no content other then whitespace), and display a count of matching lines (-c) instead of the matching lines themselves.

An advantage of this method over methods that involve piping into wc, is that you can specify multiple files and get a separate count for each file:

$ grep -cve '^\s*$' *.hh

config.hh:36
exceptions.hh:48
layer.hh:52
main.hh:39

This command count number of non-blank lines.
cat fileName | grep -v ^$ | wc -l
grep -v ^$ regular expression function is ignore blank lines.