I want to have the timestamps of the files containing a specific string in them.
I have tried using different combinations of find command and grep and ls but not giving the desired output.

find $HOME/bin/shells -name "*" -print -exec ls -Flt {} -exec grep -i "abc" '{}' \;

Please help.

When you do something like this, each time that ls runs, it will be with a single file name. So using -l to sort the output doesn't make sense.

And your -name "*" is going to match everything.

find $dir -type f -exec grep -q abc {} \; -exec ls -Fl {} \;

is the syntax that comes closest to what you're trying to do. Only if the first exec succeeds will the second exec be attempted. The order is important.

Thanks for the reply.
I am using the same command and it is funny that even though I use ls -Fl, the list is not in that order. I am trying to get the list with latest file top in the list. ls -Fl should do that. But I don't understand why it is not doing. Below is my command.

find $dir -name "*" -exec grep -q "abc" {} \; -exec ls
-Flt {} \; | awk '{print $8}' | read timestamp.

I am expecting to have timestamp = Latest file time stamp.

Please suggest. Thank you.

It's like I said, each line is output by an individual execution of ls.
ls -lt a b c
can sort the lines that are output. But
ls -lt a ; ls -lt b ; ls -lt c
cannot.

Since you're looking for every file, there's no need for the find command:

l -t `grep -lF 'abc' $HOME/bin/shells/*` | awk '{print $6,$7,$8}'

Sorry I do not understand.
Do you think I can get the latest file on the top of the list with the above command? (or if we modify it)?

Thanks for the help.

I included the -t option with the "l" command to sort by date, newest first. Give it a try and see if it does what you need. Are you only trying to print the date to the screen, or other information (such as filename) too?