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I have a file that looks like this:

foo03a
foo02b
quux01a
foo01a
foo02a
foo01b
foo03b
quux01b

I'd like it ordered by the last character (so a and b appear together) and then by the preceding number, and then by the prefix (though this is not essential). So that it results in:

foo01a
quux01a
foo02a
foo03a
foo01b
quux01b
foo02b
foo03b

It actually doesn't particularly matter where quux01a and quux01b appear, as long as they're in the relevant group -- they can appear as shown, before foo01b, or after foo03b.

Why? These are server names used in a blue/green deployment, so I want the 'A' servers together, then the 'B' servers.

I found the -k switch to GNU sort, but I don't understand how to use it to specify a particular character, counting from the end of the string.

I tried cat foos | rev | sort | rev, but that sorts foo10a and foo10b (when we count up that far) into the wrong place.

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2
  • cat foos |rev | sort |rev foo01a foo02a foo03a foo01b foo02b foo03b this is what I'm getting. Isn't it what you want ? Commented Mar 29, 2016 at 14:22
  • sort -k1.6 -k1.4,1.5n -k1.1,1.3 file Commented Mar 29, 2016 at 14:26

2 Answers 2

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2

I found a solution using GNU awk's match function:

cat foos | \
  gawk 'match($0, /([^0-9]+)([0-9]+)([^0-9]+)/, a) {print a[3], a[2], $0}' | \
  sort | cut -d' ' -f3

The gawk command uses regex captures to generate the sort key, so that I end up with the following:

a 03 foo03a
b 02 foo02b

...etc.

Run that through sort, cut for the fields I want. Done.

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3
  • awk doesn't need cat, just so you know. I'd use something like sed 's/\(.*\)\([0-9][0-9]*\)/\1\x02\2\x02/' infile | sort -t $'\002' -k3,3 -k2,2n -k1,1 | tr -d $'\002' here. Commented Mar 29, 2016 at 17:15
  • can also be done with any version of sed, using basic regular expressions: sed -e 's/\(^[^0-9]*\)\([0-9]\+\)\(.\)$/\3 \2 \1\2\3/' foos | sort | cut -d' ' -f3 Commented Mar 30, 2016 at 1:53
  • 1
    Yes, I know that awk doesn't need cat. The actual list is generated from a jq query... Commented Mar 30, 2016 at 8:44
1

You can sort by character position (indicated by number after .) of field (here field 1):

sort -k1.6 -k1.4,1.5n -k1.1,1.3 file.txt

Example:

$ cat file.txt 
foo03a
foo02b
foo01a
foo02a
foo01b
foo03b

$ sort -k1.6 -k1.4,1.5n -k1.1,1.3 file.txt 
foo01a
foo02a
foo03a
foo01b
foo02b
foo03b

Reverse way, using rev:

$ rev file.txt | sort -k1.1,1.1 -k1.2,1.3n -k1.4 | rev
foo01a
foo02a
foo03a
foo01b
foo02b
foo03b
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5
  • 1
    That's character position from the start of the string, right? I need character position from the end. I'll clarify the question. Commented Mar 29, 2016 at 14:34
  • @RogerLipscombe Yes..you just need rev then.. Commented Mar 29, 2016 at 14:36
  • @RogerLipscombe Check my edits.. Commented Mar 29, 2016 at 14:38
  • 1
    Hmm. That sorts foo10a (etc.) into the wrong place... Commented Mar 29, 2016 at 14:43
  • @RogerLipscombe Yes, it does..i think your best bet would be go with the first one (if you can).. Commented Mar 29, 2016 at 14:45

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