Bash: How to sort files in a directory by number of lines?

Question

I'm new to bash so my knowledge is pretty limited, especially when it comes to writing scripts. Is there any way in which a script can read into the directories it is given and sort the files inside in descending order, from the one with most lines to the one with least?

Answer 1

find /some/dir -type f -print0 | wc -l --files0-from=- | sort -n -r

Should do what you want

the find program scans directory /some/dir recursively, outputs the full path for each file it finds ( -type f means file as in not a directory/socket/etc). the output list uses nul-terminated strings ( -print0 ) in order to safely deal with dodgy filenames.

That list of filenames feeds into wc (wordcount) which uses ( --files0-from=- ) to expect a nul-terminated filelist as input, and for each file it prints the number of lines ( -l ) in front of the filename.

That list in turn, feeds into sort which sorts the list in reverse ( -r ) numeric ( -n ) fashion; and since the linecount is in front of the filenames, that means the longest file (most lines) is on top.