Bash - How to find the largest file in a directory

Question

We're just starting a UNIX class and are learning a variety of Bash commands. Our assignment involves performing various commands on a directory that has a number of folders under it as well.

I know how to list and count all the regular files from the root folder using:

find . -type l | wc -l

But I'd like to know where to go from there in order to find the largest file in the whole directory. I've seen somethings regarding a du command, but we haven't learned that, so in the repertoire of things we've learned I assume we need to somehow connect it to the ls -t command.

And pardon me if my 'lingo' isn't correct, I'm still getting used to it!

Answer 1

This will find the largest file or folder in your present working directory:

ls -S /path/to/folder | head -1

To find the largest file in all sub-directories:

find /path/to/folder -type f -exec ls -s {} \; | sort -nr | awk 'NR==1 { $1=""; sub(/^ /, ""); print }'

Answer 2

Try following command :

find /your/path -printf "%k %p\n" | sort -g -k 1,1 | awk '{if($1 > 500000) print $1/1024 "MB" " " $2 }' |tail -n 1 

This will print the largest file name and size and more than 500M. You can move the if($1 > 500000),and it will print the largest file in the directory.

Answer 3

du -aS /PATH/TO/folder | sort -rn | head -2 | tail -1

or

du -aS /PATH/TO/folder | sort -rn | awk 'NR==2'

Answer 4

To find the top 25 files in the current directory and its subdirectories:

find . -type f -exec ls -al {} \; | sort -nr -k5 | head -n 25

This will output the top 25 files by sorting based on the size of the files via the "sort -nr -k5" piped command.

Same but with human-readable file sizes:

find . -type f -exec ls -alh {} \; | sort -hr -k5 | head -n 25

Answer 5

On Solaris I use:

find . -type f -ls|sort -nr -k7|awk 'NR==1{print $7,$11}' #formatted

or

find . -type f -ls | sort -nrk7 | head -1 #unformatted

because anything else posted here didn't work. This will find the largest file in $PWD and subdirectories.

Answer 6

This script simplifies finding largest files for further action. I keep it in my ~/bin directory, and put ~/bin in my $PATH.

#!/usr/bin/env bash
# scriptname: above
# author: Jonathan D. Lettvin, 201401220235

# This finds files of size >= $1 (format ${count}[K|M|G|T], default 10G)
# using a reliable version-independent bash hash to relax find's -size syntax.
# Specifying size using 'T' for Terabytes is supported.
# Output size has units (K|M|G|T) in the left hand output column.

# Example:
#   ubuntu12.04$ above 1T
#   128T /proc/core

# http://stackoverflow.com/questions/1494178/how-to-define-hash-tables-in-bash
# Inspiration for hasch: thanks Adam Katz, Oct 18 2012 00:39
function hasch() { local hasch=`echo "$1" | cksum`; echo "${hasch//[!0-9]}"; }
function usage() { echo "Usage: $0 [{count}{k|K|m|M|g|G|t|T}"; exit 1; }
function arg1() {
    # Translate single arg (if present) into format usable by find.
    count=10; units=G;  # Default find -size argument to 10G.
    size=${count}${units}
    if [ -n "$1" ]; then
        for P in TT tT GG gG MM mM Kk kk; do xlat[`hasch ${P:0:1}`]="${P:1:1}"; done
        units=${xlat[`hasch ${1:(-1)}`]}; count=${1:0:(-1)}
        test -n "$units" || usage
        test -x $(echo "$count" | sed s/[0-9]//g) || usage
        if [ "$units" == "T" ]; then units="G"; let count=$count*1024; fi
        size=${count}${units}
    fi
}
function main() {
    sudo \
        find / -type f -size +$size -exec ls -lh {} \; 2>/dev/null | \
        awk '{ N=$5; fn=$9; for(i=10;i<=NF;i++){fn=fn" "$i};print N " " fn }'
}

arg1 $1
main $size