Shell Programming and Scripting

I need a unix command which will find all the files greater that a particular date in the file name.

say for example I have files like(filenaming cov : filename.YYDDMMSSSS.txt)

Code:

abc.201206015423.txt
abc.201207013456.txt
abc.201202011234.txt
abc.201201024321.txt
efg.201202011234.txt
efg.201201024321.txt

Now say I want to find abc.* files whose date is >= 20120201.
Can you please provide a unix command to get the file names ?

Thanks in advance

it is really day before month? are the files in the same directory or do you need to search sub directories?

Thanks for responding.
All the files are in same directory.
The date can be any date. I will have a variable for that.
I just gave a date in the example.

Thanks

Think this should do you:

Code:

$ awk -F'.' 'int($2)>'201206000000' {print $0}' test.sh
abc.201206015423.txt
abc.201207013456.txt

I didnt clearly get it. Why do we need test.sh here ?

You say that the file naming convention is
but the example looks a lot more like the convention is
I can't tell which it is from the data you supplied.
It is good that the convention uses YYYY instead of YY so it can handle century roll over issues like the Y2K problems some of us lived through.

If the convention is YYYYMMDDSSSS the script Vryali gave works fine (assuming that test.sh is a file containing a list of the files you're examining). If the convention is YYYYDDMMSSSS, you have more work to do to rearrange the month and day into an order that will sort correctly.

Hi
I tried this

Code:

ls abc.* | awk -F'.' 'int($2)>'201206000000' {print $0}'

and it doesnt show up anything

Thanks

---------- Post updated at 01:11 PM ---------- Previous update was at 01:09 PM ----------

sorry the naming convention is filename.YYYYMMDDSSSS.txt
still no luck.