Sort set of string elements in descending order using shell script

Question

I am trying to sort set of string elements in descending order but the result surprises me. I can see 7.1.1 comes before 7.10.1, not sure why.

$ cat sort.sh 
#!/bin/bash
for i in $( echo "5.2.0 5.1.2 7.1.1 7.10.1" | tr ' ' '\n' | sort -r )
do
  echo $i
done

Output:

$ ./sort.sh 
7.1.1
7.10.1
5.2.0
5.1.2

Expected output:

$ ./sort.sh 
7.10.1
7.1.1
5.2.0
5.1.2

Am i missing anything here? Any tips would be really helpful.

Thanks

Look at an ASCII table and where . and 0 are in it.

Shawn
– Shawn

2022-11-12 17:22:04 +00:00
Commented Nov 12, 2022 at 17:22 — Shawn
– Shawn, Commented Nov 12, 2022 at 17:22
I think you want sort -rn

sseLtaH
– sseLtaH

2022-11-12 17:32:25 +00:00
Commented Nov 12, 2022 at 17:32 — sseLtaH
– sseLtaH, Commented Nov 12, 2022 at 17:32

Jukka Matilainen · Accepted Answer · 2022-11-12 17:53:47Z

2

Sort uses the sorting rules specified by the current locale. See "Sort does not sort in normal order!" in GNU Coreutils FAQ. GNU sort (1) manual page contains this warning:

*** WARNING *** The locale specified by the environment affects sort order. Set LC_ALL=C to get the traditional sort order that uses native byte values.

If you specifically want to sort version numbers, see the -V command line option in GNU sort:

-V, --version-sort

natural sort of (version) numbers within text

edited Nov 12, 2022 at 17:53

answered Nov 12, 2022 at 17:48

Jukka Matilainen

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