Find largest file within a directory

grob115 · 09-03-2011, 03:10 AM

Am interested in writing a script that can return to me say the top 5 files in terms of size under a specific directory. The difficulty is this directory may have any number of sub-directories under it, and each of these directories may or may not have files in them. Any idea how this can be done?

yooy · 09-03-2011, 03:29 AM

you will need to go throuht all files with "find" and save biggest files as list that gets updated when bigger file is found

Nominal Animal · 09-03-2011, 03:41 AM

Use find to list all files with file size in bytes first. Then sort the output by the file size. Finally, use sed to output the N largest files (5, below), removing the file size in bytes.

Code:

find dir(s)... -type f -printf '%s %p\n' | sort -rg | sed -ne '1,5 s/^[^ ]*//p'

grob115 · 09-03-2011, 11:21 PM

Nominal Animal, thanks.

ArthurSittler · 09-04-2011, 12:42 AM

I find ls with appropriate flags works well. Pipe the output from ls through sort, as above, and head with the number of largest files you want. Then I used awk to invoke find with system() to display the complete relative path.
I suppose starting out with find is the more direct approach.

kurumi · 09-04-2011, 02:35 AM

Quote:

Originally Posted by ArthurSittler

I find ls with appropriate flags works well. Pipe the output from ls through sort, as above, and head with the number of largest files you want. Then I used awk to invoke find with system() to display the complete relative path.
I suppose starting out with find is the more direct approach.

This is not the correct approach. Using ls to list files (with ls options) spells trouble if files contains white spaces. And why would you invoke find within awk? You are making alot of overheads doing things like this.