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From: Bill Campbell <bill@celestial.com>
Subject: Re: sco-list: Locating large files
Date: Wed, 10 Jan 2001 16:52:42 GMT
References: <t5o3s4sgrqsqb6@corp.supernews.com> 


On Wed, Jan 10, 2001 at 07:30:12AM -0000, Venugopal wrote:
>Frequently I need to create space on my server.I have written the 
>following script to locate the biggest files in a given directory.
># shell script fbf to get given no. of big files
># usage fbf nn where nn is the requd. no. of files
>du -k -a | sort -n -r | head -$1 | awk -e 
>The above script gives me the names of the sub-directories and files.
>Is there a way to eliminate the directory entries?
>Is there a simpler way to do this job?

Here's a crude shell script I wrote at least 15 years ago (it still uses
awk, and everything I've written since 1990 or so is perl :-).  I call this
``lsize'', and it totals the size of all files in a directory.  The -r
option sorts by size in reverse order.

INTERNET:   bill@Celestial.COM  Bill Campbell; Celestial Software LLC
UUCP:               camco!bill  PO Box 820; 6641 E. Mercer Way
FAX:            (206) 232-9186  Mercer Island, WA 98040-0820; (206) 236-1676
URL: http://www.celestial.com/

``Never blame a legislative body for not doing something.  When they do
nothing, that don't hurt anybody.  When they do something is when they
become dangerous.''
    Will Rogers


case $# in
        0)      args="*"        ;;
        *)      case $1 in
                -r)     SORT=TRUE;

case $SORT in
        TRUE) ls -li $* | sort +0n -u |
                sort +5rn |awk '{sum+=$6; printf("%8d %8d %s\n",sum, $6, $10)}' 
                ls -l $* | awk '{sum+=$5; printf("%8d %8d %s\n",sum, $5, $9)}'


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