Shell Programming and Scripting

You don't need the int() for this comparison, and it will hurt you if you're not using an awk that uses 64-bit ints. Just use:

Code:

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

Hi Don
Thanks for the reply. That worked.
But one small change in my question. Sorry for not letting you know this earlier.
My file names are

Code:

abc.C201206015423.txt
abc.C201207013456.txt
abc.C201202011234.txt
abc.C201201024321.txt
efg.C201202011234.txt
efg.C201201024321.txt

Code:

ls abc.* | awk -F'.' '$2>'C201206000000' {print $0}'

above didnt work.

Thanks

Not surprising. With a numeric string 201206000000, the value being compared is 201206000000. With the string C201206000000, awk sees a variable name and since the variable hasn't been set, the value being compared is 0. You need to compare a string to a string, so try:

Code:

ls abc.* | awk -F'.' '$2>"C201206000000" {print $0}'

Presumably at some point you'll want to put this in a script where you have the start date in a shell variable (e.g., STARTDATE). When you get there, you'll probably want something like:

Code:

STARTDATE=20120600000000
ls *.*.txt | awk -F. '$2 > "C'$STARTDATE'" {print}'

What shell do you use? Do you want a script that'd list these certain files, or a function you'd use from the command line?

Hi
I need a script and it is c shell.