Hello! I'm wondering if some one could help me with the following with a UNIX shell scripting. a) I want to count the files according to the date. Please find the details as follows, Oct 15, 2009 - 3 files Oct 14, 2009 - 2 files ......... ......... Oct 14, 2008 - 4 files ........ ........ I just need the files count on a particular day and the date should indicate the "Month Day, Year - No. of files" in the result. I tried, but i'm not able to get the year in the result. Your help would be greatly appreciated. Thanks

How about sharing what you have, and we'll help you from there?

Hi Nails, please find this. ls -lt | grep "^-" | awk '{Date_count=$6" "$7; freq[Date_count]++}END {for (date in freq) printf "%s\t%d\n", date, freq[date] }' i have this , but it's not giving me what i want. Please help me with this. Thanks!!

You did the hard part. Remember that on legacy Unix systems that the ls command for the first 6 months shows the time instead of the year. (GNU ls has an option to show the year). Therefore you may have trouble with this script in the 4th quarter: ls -lt | grep "^-" | awk '{ if($8 ~ /:/) $8=2009 Date_count=$6" "$7", "$8 freq[Date_count]++} END {for (date in freq) printf "%s\t%d\n", date, freq[date] }'

Hi Nails! This is great solution for my issue. I have tried your modified script and i get results as i expected even though the year doesn't show up for the first 6 months, and i'm happy for that. Thank you.