Sorting date field in unix
bash, shell, sorting, unix
Solution
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.
Problem
I have text file which contains hundreds of thousands of records. One of the fields is a date field. Is there is any way to sort the file based on the date field? ``` 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 ``` The Output should come as ``` 19-MAR-12 03.53.38.189606000 PM 19-MAR-12 03.54.32.595348000 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 27-MAR-12 10.28.14.797580000 AM 27-MAR-12 07.28.02.828746000 PM 28-MAR-12 12.28.02.652969000 AM 09-APR-12 04.08.43.632279000 AM ``` I have tried the sort command to order the date (taking the date field as a string), but it is not giving the correct output.