Hi there, can anybody tell me how to compare the size of the given files( 2 or more) and then display the one with bigger size. ( all this is in command line, I don't have GUI)

Create the following script. Make it executable ( chmod ugo+x scriptname ). Execute it with the names of the two files you want to compare. #!/bin/sh if [ $# -ne 2 ] ; then echo "USAGE: $0 file1 file2" exit 4 fi file1=$1 file2=$2 file1size=`ls -l $file1 | awk '{ print $5 }'` file2size=`ls -l $file2 | awk '{ print $5 }'` if [ $file1size -eq $file2size ] ; then echo "$file1 and $file2 are the same size" elif [ $file1size -gt $file2size ] ; then echo "$file1 is larger with $file1size bytes" else echo "$file2 is larger with $file2size bytes" fi

thanks for the script. ok, this script would work fine with two files. what if in a given directory there are several files and I just want to display the size of the biggest file. I figured out a way to display the file size and the file name, but all i want is the file size and not the file name. here is the command: ls -s | sort -nr | head -1 cannot figure out how can I just display the file size.

ls -s | sort -nr | head -1 | awk '{ print $1 }' Bear in mind that the -s flag shows a block count not a byte count.

thanx david. so how do I show a byte count?

Don't forget to ignore folder's size: find ./ -type f -maxdepth 1 | xargs ls -lk | sort -n -k 5 | tail -1 | awk '{print $5*1024,"bytes"}' Modify awk's output as you wish.

thanx man, that would work.