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Sort by dd/mm/yy format

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# 1  
Old 02-03-2011
Sort by dd/mm/yy format
i tried this
Code:
awk '!a[$1$2]++' filename

to sort my returned list. But, I don't think I'm using the right thing to sort this list.

Code:
01/Feb/2011   -- User Count : 27
  31/Jan/2011   --  User Count : 21
  02/Feb/2011   -- User Count : 24
  30/Jan/2011   --  User Count : 4

I'm trying to get the output to look like this:
Code:
30/Jan/2011   --  User Count : 4
31/Jan/2011   --  User Count : 21
01/Feb/2011   -- User Count : 27
 02/Feb/2011   -- User Count : 24

# 2  
Old 02-03-2011
We should be able to use "sort". This assumes that the date dd/mmm/yy starts in column 1 . We also assume normal locale settings.
The key part of this "sort" command is the "M" (Compare as months). See "man sort".

Code:
cat filename | sort -t '/' -k 3.1n,3.2 -k 2.1M,2.3 -k 1.1n,1.2

This User Gave Thanks to methyl For This Post:
alten
# 3  
Old 02-03-2011
TY, that works nicely
# 4  
Old 03-13-2011
Good job.

Code:
cat filename | sort -t '/' -k 3.1n,3.2 -k 2.1M,2.3 -k 1.1n,1.2

Helped me heaps (with a little tweaking for my different need). Thanks.
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