How to sort the output of "grep -l" chronologically by newest modification date last?

Question

I'm using the -l flag with grep to just print the matching file names.

I want to then list the results (files) in ascending order i.e. newest last.

Obviously

grep -l <pattern> *.txt | ls -rt

is not what I need, since the ls -lrt merely outputs all files.

Answer 1

Try:

ls -rt *.txt | xargs grep -l <pattern>

We first use ls to list *.txt files and sort them by modification time (newest last), then for each entry run them through grep so we only print out files that contain the pattern.

Answer 2

Here is another way

grep -l <pattern> *.txt | xargs ls -ltr --time-style="+%Y%m%d_%H%M"

For example

$ grep -l argparse * | xargs ls -ltr --time-style="+%Y%m%d_%H%M"
-rwxr-xr-x 1 raju 197121   979 20171030_2323 list_dep.py
-rwxr-xr-x 1 raju 197121 14329 20180129_2216 grep_installed.py
-rwxr-xr-x 1 raju 197121  7517 20180129_2216 popsort.py

To just list the files without timestamps etc.,

grep -l <pattern> * | xargs ls -1tr

For example

$ grep -l argparse * | xargs ls -1tr
list_dep.py
grep_installed.py
popsort.py

Answer 3

I know this is an old question but thought of adding my suggestion for someone who is searching for the same..

for i in $(ls -rt *.txt); do grep <pattern> $i; done