Sorting based on File name

Shell Programming and Scripting
1

Hi All

I have a requirement to list all the files in chronological order based on the date value in the file name.For ex if I have three files as given below

ABC_TEST_20160103_1012.txt
ABC_TEST_20160229_1112.txt
ABC_TEST_20160229_1112.txt

I have written code as given below to list out the files.This is the command which I am trying to execute from Data stage job which is running on UNIX OS.

ls -m ABC*|sort -t_ -k3

The above code is working in this scenario.But in case if we receive a file name as below, this will not work, I need to change the command accordingly,as date is coming as first portion of the file name.

20160122_2130_ABC_TEST.txt

I want to write a common command in such a way that it will sort automatically based on the date value in the file name, irrespective of the position where the date value is coming in the file

2

Try something like:

ls | sed 's/.*\([0-9]\{8\}\)/\1 &/' | sort -k1,1n | cut -c10-

This assumes that 8 digits always constitute a date..

--
If the date is always followed by an underscore you could make this a bit more precise like so:

ls | sed 's/.*\([0-9]\{8\}\)_/\1 &/' | sort -k1,1n | cut -c10-
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3

And if the filenames always contain an 8 digit date followed by an underscore and a 4 digit time ( YYYYMMDD_hhmm ) and the desire is to sort on the date and time, you could try:

ls | sed 's/.*\([0-9]\{8\}_[0-9]\{4\}\)/\1 &/' | sort -k1,1 | cut -c14-

Note that I removed the sort -n option because an alphanumeric sort will get the desired results on the 1st pass while a numeric sort would stop comparing at the underscore (ignoring the timestamp on the 1st pass).

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