Korn: How to zero fill df output so it will sort properly

shew01 · #1 (**permalink**) 06-04-2008

I'm looking for a way in Korn shell to zero fill (or space fill) the output from df so that it will sort properly.

"Raw" output from df -k:

Code:

df -k

Filesystem            kbytes    used   avail capacity  Mounted on
/dev/vx/dsk/rootvol  4131866 3593302  497246    88%    /
/proc                      0       0       0     0%    /proc
fd                         0       0       0     0%    /dev/fd
mnttab                     0       0       0     0%    /etc/mnttab
swap                 19891520     144 19891376     1%    /var/run
swap                 19913552   22176 19891376     1%    /tmp
/dev/vx/dsk/u01      112302757 30833295 80346435    28%    /u01
/dev/vx/dsk/node@1     96975    4977   82301     6%    /global/.devices/node@1
/dev/vx/dsk/node@2     96975    4988   82290     6%    /global/.devices/node@2
/dev/vx/dsk/stkdg/s03
                     1047527424 60483821 925353439     7%    /s03
/dev/vx/dsk/stkdg/s02
                     1048576000 266754165 732957997    27%    /s02
/dev/vx/dsk/sandg/s01
                     355923968 322572197 31422792    92%    /global/s01

Sorted df output (which incorrectly sorts 28% and 27% after 7%):

Code:

df -k | grep / | awk '{ print $5}' | sort -r

92%
88%
7%
6%
6%
28%
27%
1%
1%
0%
0%
0%

I would like the output to display in the following format:

Code:

92%
88%
28%
27%
 7%
 6%
 6%
 1%
 1%
 0%
 0%
 0%

Any ideas?

System Shock · #2 (**permalink**) 06-04-2008

The output you are getting is not "incorrect". It is actually correct with the flags you are giving sort.
You have to tell sort that you are sorting numbers and not text.

Code:

:> df -k | grep / | awk '{print $5}' | sort -gr
100%
7%
2%
0%
0%

A quick look at the man pages would've given you the answer.

vgersh99 · #3 (**permalink**) 06-04-2008

Code:

df -k | awk '/[/]/ { print $5}' | sort -rn

radoulov · #4 (**permalink**) 06-04-2008

Or:
if perl is acceptable:

Code:

df -k | 
  perl -ane'
    /%/ and push @x,$F[4];
    print join "\n" , reverse sort { $a <=> $b } @x 
if eof'

Otherwise:

Code:

df -k |awk '1<NR&&NF>1{print $5}'|sort -nr

shew01 · #5 (**permalink**) 06-04-2008

System Shock,

>>A quick look at the man pages would've given you the answer.

I don't think so:

Code:

df -k | grep / | awk '{print $5}' | sort -gr
sort: illegal option -- g
usage: sort [-cmu] [-o output] [-T directory] [-S mem] [-z recsz]
        [-dfiMnr] [-b] [-t char] [-k keydef] [+pos1 [-pos2]] files...

shew01 · #6 (**permalink**) 06-04-2008

vgersh99,

Quote:

Originally Posted by vgersh99

Code:

df -k | awk '/[/]/ { print $5}' | sort -rn

Cool! This seems to work. Thanks for the help.

radoulov · #7 (**permalink**) 06-04-2008

And another one:

Code:

df -k | 
  while read fs kb us av ca mo; do 
    case $ca in 
      *[0-9]% ) printf "%s\n" "$ca";;
    esac
done | sort -rn

You should use df -P on Linux.

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