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    • Alan KilbornA
      Alan Kilborn @PeterJones
      last edited by

      @PeterJones said in Wildcard in replace field:

      That would be nice.

      It would probably also be something a dev would have to remember to do by hand when it changes, instead of being automated, so it might be a poor idea anyway (I don’t trust the hoomans to remember to update these types of things).

      1 Reply Last reply Reply Quote 1
      • guy038G
        guy038
        last edited by guy038

        Hi, @alan-kilborn, @peterjones, @ekopalypse, and All,

        Alan and Peter, thank you so much for these clarifications ;-)

        Although the last Boost regex 1.7.3 version is not used and that Notepad++ is build with Boost v1.7.0 since N++ v7.8.0, the differences between the search regex documentation of Boost versions v1.5.5 and v1.7.0, mentioned in my previous post, are identical and are still valid ;-))


        So, the good news is that our new regex library does support the Perl’s backtracking control verbs. Woooow !! I’m quite eager to investigate this new side of regular expressions:-)) I will keep you informed of the interest of these new functions and will try to provide you with relevant examples for a better understanding…

        As a result, I have also updated the FAQ contents !


        Now, while browsing the Boost C++ libraries website, I came across a thorny problem : when we speak of Boost v1.70.0, I think that it’s an abuse of language ! Because the strings 1.55.0 and 1.70.0 refer to a documentation version and NOT to a library version :-( Indeed :

        • Formerly, we were using the Boost-Regex library from the Boost-1.54.0 version of the Boost C++ libraries and we referred to its Boost 1.55.0 library documentation

        • Since Notepad v7.7, we’re using the Boost-Regex-5.1.3 library from the Boost-1.64.0 version of the Boost C++ libraries and we refer to its Boost 1.70.0 library documentation

        Refer to all these links, below. It’s a bit tricky !? Compare the different links, two by two

        https://www.boost.org/doc/libs/1_55_0/

        https://www.boost.org/doc/libs/1_70_0/

        https://www.boost.org/doc/libs/1_55_0/libs/regex/doc/html/index.html

        https://www.boost.org/doc/libs/1_70_0/libs/regex/doc/html/index.html

        https://www.boost.org/doc/libs/1_55_0/libs/regex/doc/html/boost_regex/background_information/history.html

        https://www.boost.org/doc/libs/1_70_0/libs/regex/doc/html/boost_regex/background/history.html


        Could you confirm my guesses ? TIA !

        Cheers,

        guy038

        P.S :

        The last version uses the Boost-Regex-5.1.4 library from the Boost-1.72.0 version of the Boost C++ libraries and refers to the Boost 1.73.0 library documentation

        See :

        https://www.boost.org/doc/libs/1_73_0/

        https://www.boost.org/doc/libs/1_73_0/libs/regex/doc/html/index.html

        https://www.boost.org/doc/libs/1_73_0/libs/regex/doc/html/boost_regex/background/history.html

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        • daxliniereD
          daxliniere
          last edited by

          A big thank you to everyone who replied to my question offering help. @Alan-Kilborn’s suggestion seems to work perfectly.

          All the best to this great community!
          -Dax.

          1 Reply Last reply Reply Quote 2
          • 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( |$))

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                          • 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([ ,'!\.\?"”\)]|$))

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