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    • daxliniereD
      daxliniere
      last edited by

      Hey all!
      I have no idea why, but for some reason, this solution has stopped working…
      Seriosuly, I have no idea what’s going on, but (\d\d:\d\d):(?=\d\d) no longer finds a string of text that looks like 0:10:02 using RegEx.

      Is it possible a recent update to NPP might have broken this or changed syntax??
      I’m so confused.

      Thanks in advance!
      -Dax.

      astrosofistaA 1 Reply Last reply Reply Quote 0
      • astrosofistaA
        astrosofista @daxliniere
        last edited by

        Hi @daxliniere

        The quoted regex matches strings with two leading numbers, as “00:12:12”.
        In order to match strings with only one leading number, just remove a “\d” from the expression, as (\d:\d\d):(?=\d\d).

        Have fun!

        daxliniereD 1 Reply Last reply Reply Quote 2
        • daxliniereD
          daxliniere @astrosofista
          last edited by

          @astrosofista Thank you so very much!! It works exactly as before.
          I have no clue what could have changed, but this has solved the problem.

          I really appreciate your time on this. I was in the middle of a job and got stuck on this. Was comtemplating manual correction of ~400 lines!

          Have a great weekend. :)
          -Dax.

          1 Reply Last reply Reply Quote 1
          • daxliniereD
            daxliniere
            last edited by

            AHHH!!! I understand it now!

            The data I had been receiving previously had trailing zeros so every line had xx:xx:xx formatting. It seems whoever was typing this started dropping the trailing zeros.
            Is there a more robust form of (\d:\d\d):(?=\d\d) needed?

            PeterJonesP 1 Reply Last reply Reply Quote 1
            • PeterJonesP
              PeterJones @daxliniere
              last edited by

              @daxliniere ,

              You say “trailing”, but the regex shown (as modified by @astrosofista) has removed the requirement for leading zeros. To make it match both, I’d do (\d{1,2}:\d\d):(?=\d\d), which will match whether there is one digit or two.

              But if you give an example of “with” or “without” “trailing zeros”, we can be more confident of what you’re really asking for.

              Thomas 2020T 1 Reply Last reply Reply Quote 2
              • Thomas 2020T
                Thomas 2020 @PeterJones
                last edited by

                @PeterJones
                instead:
                (\d{1,2}:\d\d):(?=\d\d)

                maybe that’s better?
                (:\d{1,2}):

                the shorter the pattern, the better

                PeterJonesP 1 Reply Last reply Reply Quote 0
                • PeterJonesP
                  PeterJones @Thomas 2020
                  last edited by

                  @Pan-Jan ,

                  Your regex does not provide all the same groups (and thus features) of the regex developed through this thread, and thus likely misses some of the functionality needed

                  1 Reply Last reply Reply Quote 0
                  • Thomas 2020T
                    Thomas 2020
                    last edited by Thomas 2020

                    @PeterJones said in Wildcard in replace field:

                    and thus likely misses some of the functionality needed

                    show on the example in this thread that it doesn’t work

                    (\d{1,2}:\d\d):(?=\d\d)
                    but this one has disadvantages, and he will also mark it here
                    126:23:45
                    126:23:450
                    12:23:45:67

                    it will be correct

                    (^| )(\d{1,2}:\d\d):(?=\d\d( |$))

                    1 Reply Last reply Reply Quote 0
                    • Thomas 2020T
                      Thomas 2020
                      last edited by Thomas 2020

                      This post is deleted!
                      1 Reply Last reply Reply Quote 0
                      • Thomas 2020T
                        Thomas 2020
                        last edited by Thomas 2020

                        (^| )(\d{1,2}:\d\d):(?=\d\d([ ,'!\.\?"”\)]|$))

                        1 Reply Last reply Reply Quote 0
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