Sorting date field in unix

Chronicle's solution is close, but misses the AM/PM distinction, sorting 27-MAR-12 07.28.02.828746000 PM before 27-MAR-12 10.28.14.797580000 AM. This can be modified:

sort -t- -k 3.1,3.2 -k 2M -k 1n -k 3.23,3.24

But that is still very fragile. It would be much better to convert the dates to an epoch time and compare numerically.

Try this :

Input.txt

09-APR-12 04.08.43.632279000 AM 19-MAR-12 03.53.38.189606000 PM 19-MAR-12 03.56.27.933365000 PM 19-MAR-12 04.00.13.387316000 PM 19-MAR-12 04.04.45.168361000 PM 19-MAR-12 03.54.32.595348000 PM 27-MAR-12 10.28.14.797580000 AM 28-MAR-12 12.28.02.652969000 AM 27-MAR-12 07.28.02.828746000 PM 

Code

 sort -t "-"  -k 3 -k 2M -nk 1 Input.txt

Output

19-MAR-12 03.53.38.189606000 PM19-MAR-12 03.54.32.595348000 PM19-MAR-12 03.56.27.933365000 PM19-MAR-12 04.00.13.387316000 PM19-MAR-12 04.04.45.168361000 PM27-MAR-12 07.28.02.828746000 PM27-MAR-12 10.28.14.797580000 AM28-MAR-12 12.28.02.652969000 AM09-APR-12 04.08.43.632279000 AM

This script sorts by Epoch time with nanosecond resolution:

awk '{  t = gensub(/\.([0-9]{2})\./, ":\\1:", 1, $0);  command = "date +%s%N -d \x022" t "\x022";  command | getline t;  close(command);  print t, $0;}' unsorted.txt | sort -n -k 1 | cut -d ' ' -f 2- > sorted.txt