Possible Duplicate:
How to name a file in the deepest level of a directory tree
How do I get this script to work? The goal is to find the directory at the deepest level in the directory tree and print the name of a file, a random file, maybe the first.
find -type d | declare COUNT=-1; declare P=""; while read LINE ;
do
echo $LINE
declare C
C=$(echo $LINE | cut -c3- | sed "s/\//\n\//g" | grep '/' -c)
echo $C
if [ $COUNT -gt $C ]
then
let COUNT=$C
let P=$LINE
echo "Done"
fi
done
Better solutions are also welcome but they should be one-liners.
Will not do it again.
– Robert Sundström Dec 18 '11 at 03:09