compare file sizes

joli · #1 (**permalink**) 02-04-2005

Is there a command that will return the name of the largest file within a directory? If so, can I set the returned filename into a variable?

encrypted · #2 (**permalink**) 02-04-2005

Try
LARGEST=`du –s /<path_to_dir>/* | sort –rn | cut –f2 | head –1`

thumsup9 · #3 (**permalink**) 02-04-2005

try this:

#!/bin/ksh
#say tmp is your directory
cd /tmp
A='find /tmp -type f -print | xargs ls -l | sort -r -n -k 5,5 | head -1'
echo '$A'

bhargav · #4 (**permalink**) 02-04-2005

Code:

ls -l | tr -s " " | sort -t" " +4 -n | tail -1 | cut -d" " -f 9

vgersh99 · #5 (**permalink**) 02-04-2005

'du -s' - includes directories
the 'find' solution above is recursive

largest FILE in the CURRENT directory
find . ! -name . -prune -type f -ls | sort -r -k 7,7 | head -1

Assuming the file name does contain embedded space [in ksh]:
A=$(find . ! -name . -prune -type f -ls | sort -r -k 7,7 | nawk 'FNR==1{print $NF; exit}')

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