How can I combine this into a one-liner — bash

Question

So I want to recursively search a directory with thousands of folders for a specific file extension and get the date it was added then filter by a specific year. The code below works but is there a way it can be converted to a one-liner?

1. get all files ending with xml & write to tmp.txt

    find . -name "*xml" >> tmp.txt 

2. get permission info & timestamp info for all those files in tmp.txt

    foreach i ( ` cat tmp.txt ` ) ls -ltrh $i >> tmp2.txt end 

3. column 8 contains dates as years, so get all files with years greater than a specific 2014...

    awk '$8 >=2014' tmp2.txt >> tmp3.txt 

Answer 1

Try this one:

find . -type f -name '*xml' -newermt "2014-01-01" -ls

From man find :

   -newerXY reference
          Succeeds if timestamp X of the file being considered is newer
          than timestamp Y of the file reference.   The letters X and Y
          can be any of the following letters:

          a   The access time of the file reference
          B   The birth time of the file reference
          c   The inode status change time of reference
          m   The modification time of the file reference
          t   reference is interpreted directly as a time

Answer 2

What I would do in a bash shell :

find . -mtime -$((365*4)) -name "*xml" >> tmp.txt  

This is just an example, adapt the math to your needs...

Another solution :

find . -name '*xml' -printf '%TY %p\n' | awk '$1 >= 2014 {$1=""; print}'