Using regular expression to search for a specific pattern in UNIX

Question

I have file names like ABCD20140207090842 ABCD20140207090847 ABCD20140207090849 ABCD20140207090850 ABCD2014556644219268 ABCD20140508525691 tf in my directory.

I want to search for files with specific pattern. ie FileNameYearMonthDayHourMinSec.txt

Note: files tf and ABCD2014556644219268 should not get matched.

Answer with exact pattern would be appreciated.

Answer 1

Based on NAME, followed by DATE.txt I get this:

find . -regex ".*[19|20][0-9][0-9][0-1][0-2][0-3][0-9][0-2][0-9][0-6][0-9][0-6][0-9]\.txt$"

This doesn't account for leap-years though, and could match dates such as 31 Feb.

Answer 2

Change the idea. Here is the script no matter leap years or not.

Using GNU date to identify the date

for file in *.txt
do
  time=${file%.*}         # remove suffix and get ABCD20140207090842 
  time=${time:(-14)}      # get the date/time 20140207090842
  time="${time:0:4}/${time:4:2}/${time:6:2} ${time:8:2}:${time:10:2}:${time:12}"      # convert time to: 2014/02/07 09:08:42
  date -d "$time" >/dev/null 2>&1 && echo $file
done

ABCD20140207090842.txt
ABCD20140207090847.txt
ABCD20140207090849.txt
ABCD20140207090850.txt