Hi All,
I am involved in a Datawarehousing project where my data files(ASCII text) are loaded on a Unix directory. When I do a Ls -l in Unix prompt i get a file listing along with the timestamp when these files were created/put on the server. I need to retrieve the timestamp associated with each file. Is there any command that takes in the filename as an argument and returns the timestamp associated with it?? Or is there any other way to get the timestamp for each file..
Urgent help is sought!!
Thanks in advance..
Nilotpal.

"ls -l" is pretty much it. I assume that you know that "ls -l" can indeed take arguments? So:
ls -l filename
would seem to do what you want. If this is not the case, please tell us why.

If you wanted to create a separate file that just had the filename in one column and the 'timestamp' in another...then you'd just use the ls -l command and cut the columns that you want (filename and date time....)

for example if all you files were data_xxx (where xxx differs but they all start with data_)

ls -l data_* | cut -c42-100 > my_file

Ideally with cut you'd use the -f for fields with the -d for delimiter, however the delimiter is space and the different columns have a different number of spaces as a delimiter. So you can just cut the characters 42-100 (note the 42 may differ so you need to find where the first character you want is..and 100 just goes way past the end to ensure you get all long filenames.)

There may be a way to use the -d and -f if you can specify the treat consecutive delimiters as one, however then you'd have to watch for filenames with spaces in them anyway.

Anyone know if there is a way with cut to treat consecutive delimiters as one?

__________________
Pete