3

A command outputs this:

file_0
file_1
file_10
file_11
file_12
file_13
file_14
file_15
file_2
file_3
file_4
file_5
file_6
file_7
file_8
file_9

How can I use awk or some other posix tool to actually sort it by the contiguous digits as a single number:

file_0
file_1
file_2
file_3
file_4
file_5
file_6
file_7
file_8
file_9
file_10
file_11
file_12
file_13
file_14
file_15

In general it should also work in case the digits is inside the file name, e.g.:

file_0.txt
file_1.txt
file_10.txt
file_11.txt
file_12.txt
file_13.txt
file_14.txt
file_15.txt
file_2.txt
file_3.txt
file_4.txt
file_5.txt
file_6.txt
file_7.txt
file_8.txt
file_9.txt
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  • it's not in posix sort but both GNU and FreeBSD's sort support the -V or --version-sort option, which is a "natural sort" of numbers embedded within text. – cas Sep 22 at 3:57
5
sort -nt '_' -k2

Output:

file_0
file_1
file_2
file_3
file_4
file_5
file_6
file_7
file_8
file_9
file_10
file_11
file_12
file_13
file_14
file_15

or:

file_0.txt
file_1.txt
file_2.txt
file_3.txt
file_4.txt
file_5.txt
file_6.txt
file_7.txt
file_8.txt
file_9.txt
file_10.txt
file_11.txt
file_12.txt
file_13.txt
file_14.txt
file_15.txt

Tested with FreeBSD and GNU coreutils implementations of sort but would not work with busybox implementation. All options used are specified by POSIX.

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  • I see that option in the man page now! It was not clear at all to me at first. – jsb Sep 21 at 21:08
2

Please try this:

output | awk '{print gensub("[^0-9]*","","g")  " " $0 }' | sort -n | awk '{$1=""; print $0}' | sed 's/^ //g'

This isn't the most elegant solution but it works.

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1

The answer(one of them, I'm sure) is:

sort -t _ -k 2 -g [filename with names+numbers or piped from another command with | - both situations will work ]
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