Count number of lines in files then compare which has more (BASH)

Question

I need to count the number of lines in x files and compare which has more.

The one I've done only takes two files and compares them. Any idea how to make it x amount of files?

echo Enter the filename read fn

echo Enter another file read fn1

for WORD in $(cat $fn) do
        echo "$WORD" done | wc -l

for WORD in $(cat $fn1) do
        echo "$WORD" done | wc -l

if (cat $fn | wc -l > cat $fn1 | wc -l) then
        echo First file has more lines than second file else if (cat $fn1 | wc -l > cat $fn | wc -l) then
        echo Second file has more lines than first file.

fi

Answer 1

find . -name "*.txt" -exec wc -l '{}' \; | sort -n

you can learn line size then sort them with a one liner.

Answer 2

find . -name '*txt' | xargs wc -l | sort -n

On my machine, this was faster than the -exec version.

Answer 3

wc -l * | head -n -1 | sort | tail -1 | cut -d ' ' -f 3

This will give you the filename of the file with the highest number of lines

Answer 4

Just name all files you want to compare, and sort by size (-numeric):

wc -l a.html b.html c.html | sort -n