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    Remove unwanted CRLF in paragraphs

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    • Jeroen BorgmanJ
      Jeroen Borgman
      last edited by Jeroen Borgman

      I want to change:

      Just some text<CRLF>
      I wrote to<CRLF>
      serve as example.<CRLF>
      <CRLF>
      

      into:

      Just some text I wrote to serve as example.<CRLF>
      <CRLF>
      

      In some 200 files, each files containing lots of these paragraphs.
      How can this be done?

      EkopalypseE 1 Reply Last reply Reply Quote 0
      • EkopalypseE
        Ekopalypse @Jeroen Borgman
        last edited by

        @Jeroen-Borgman

        How can this be done?

        By finding a pattern which all have in common.
        Without such a pattern, … by manually joining the lines.

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

          Hello, @jeroen-borgman, and All,

          Not difficult with regexes. Indeed !

          However I advice you to get rid of any trailing space characters, first ! Two possibilities :

          • Use the N++ option Edit > Blank Operations > Trim Trailing Space

          • Execute the regex S/R :

            • SEARCH \h+$

            • REPLACE Leave EMPTY


          Now, given, for instance, the sample text, below, without any trailing space :

          Just some text
          I wrote to<
          serve as example
          
          Just some text
          I wrote to<
          serve as example
          
          
          A single line !
          
          
          
          A last
          paragraph to
          see if the
          result is
          OK
          
          

          Then, the regex S/R :

          • SEARCH (?-s).\K\R(?!\R)

          • REPLACE \x20

          with a click on the Replace All button, would return the text :

          Just some text I wrote to< serve as example
          
          Just some text I wrote to< serve as example
          
          
          A single line !
          
          
          
          A last paragraph to see if the result is OK
          
          

          Notes :

          • First, the part (?-s) means that any dot (. ) will represents a single standard character, only and not EOL ones

          • Then the part .\K\R looks for any standard char, right before an end of line and, due to the \K syntax, the regex engine considers the regex at the right of \K, i.e. the syntax \R which represents any form of EOL ( \r\n if Windows, \n if Unix or \r if Mac )

          • Finally, the (?!\R) part is a negative look-around, i.e. a condition which must be verified. This condition force the replaceent of the EOL character(s) of a line ONLY IF it is not followed, itself, with other EOL character(s)

          • In replacement, the \x20 syntax is a synonym of the space character

          Best Regards,

          guy038

          Jeroen BorgmanJ 1 Reply Last reply Reply Quote 2
          • Jeroen BorgmanJ
            Jeroen Borgman @guy038
            last edited by

            @guy038 Thanks for this, works like a charm!
            I will study the search syntax with the REGEX doc on the side to do this myself next time.

            1 Reply Last reply Reply Quote 1
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