linux sorting of lists using the last parts using sort command [duplicate]

Question

how do I sort lists like the one below using sort or any related commands?

12.ale.panal.sel.blr.teta.hf.ew.rr.lwq.ors.2018
34.ev.we.ars.lmn.2017
5.lam.bere.meto.belagn.hede.we.e.2020
54.arad.met.kal.sil.tek.br.yz.2005

I want to sort them according to the last word which is the year chronologically?

N.B basically something like sort -t. -k-1 although -1 is not allowed. Any command just to address the last part?

the third word meaning the word after 3 dots from the start of the line — Sol, Jun 8, 2020 at 22:52
not really because they are not columns rather just words and numbers just connected with dots — Sol, Jun 8, 2020 at 22:54
You just have to tell sort that your field separator is the dot. Check the -t flag. — Eduardo Trápani, Jun 8, 2020 at 22:56
You know, one of the conditions of asking here is to do some research first. Sometimes people will give you an answer, but usually you have to show that you've tried. The command sort sorts based on fields (see the answer linked above). By default the field separator is whitespace, with -t. the field separator will be a dot. You' most likely can pick it up from there. — Eduardo Trápani, Jun 8, 2020 at 23:00

glenn jackman · Accepted Answer · 2020-06-09 12:54:37Z

1

... | awk -F. -v OFS=. '{print $NF, $0}' | sort -t. -n -k1,1 | cut -d. -f2-

where ... is whatever command you use to produce the list.

answered Jun 9, 2020 at 0:21

glenn jackman

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