How to replace a whole line starting with a given expression

  • Hello
    I would like to remove all lines in a NotePad++ text file starting with a given word.
    These lines may vary in the number of caracters. Exemple : I want to remove all lines starting with

    X-Gmail-Labels: (and then, the length of the line as well as its content may vary)
    Such as :
    X-Gmail-Labels: =?UTF-8?Q?Messages_envoy=C3=A9s,Bo=C3=AEte_de_r=C3=A9ception?=

    What to I write in the Search box ?
    Thanks !

  • @Rachad-Antonius

    by using regular expression and


    in find what, keep replace with empty you can replace all lines.

    For details about the regex see here.


  • @Claudia-Frank

    Ok, Claudia, this time you have really lost me. I don’t understand your solution! :-(

    Here’s mine:

    Find-what zone: (?-s)^X-Gmail-Labels:.*\R
    Replace-with zone: make sure this is EMPTY
    Search-mode: Regular Expression
    Wrap-around checkbox: ticked
    Action: Press Replace-All button

    That is tailored to your example, Rachad. Obviously(?) the general solution is:

    Find-what zone: (?-s)^givenword.*\R
    (all else the same)

  • @Scott-Sumner

    Hi Scott,
    I’m puzzled too, where is the circumflex and why is there a forwardslash instead of a backslash??

    I thought I wrote


    I added the \s just in case it is in front of GIVEN_WORD - might be not exactly what OP is asking for.

    Thank you for pointing this out.


  • Hello @rachad-antonius, @claudia-frank, @Scott-sumner, and All,

    Claudia, don’t forget that the \s regular syntax matches any single character of the range [\t\n\x0B\f\r\x20\x85\xA0\x{2028}\x{2029}]

    So let’s imagine this initial text, below :

    • Which begins with 3 empty lines

    • With 2 other empty lines, after line bbbbbbb

    • And the second to last line contains, ONLY the expression GIVEN_WORD

    GIVEN_WORD aaaaaaaaaaaa
    GIVEN_WORD ccccccccc
    GIVEN_WORD fffffff
    GIVEN_WORD iiiiiiiiiiiiiiiiiiiiiiiiiiiii

    With your regex, ^\s*GIVEN_WORD.*, this text turns into :


    Whereas the Scott’s regex, (?-s)^GIVEN_WORD.*\R gives :


    This later syntax seems more correct as it matches a single line, at a time ;-)) doesn’t it ? Claudia, you may click several times on the couple Find Next | Replace, with the View > Show Symbol > Show All Characters option set !

    BTW, @rachad-antonius, the \R syntax, at the end of the regex, matches the line-break characters of the current line, whatever the kind of file ( Unix, Windows, or Mac ) !

    Additional Notes :

    • In case, @rachad-antonius, that you would like, in addition, to delete entire lines, having some blank characters ( spaces and/or tabulations ), before the GIVEN_WORD, you could use the regex ^\h*GIVEN_WORD(?-s).*\R

    • Secondly , depending if you prefer the GIVEN_WORD to be matched with that exact case, or not, you could choose :

      • The regex ^\h*(?-i)GIVEN_WORD(?-s).*\R for an exact case match

      • The regex ^\h*(?i)GIVEN_WORD(?-s).*\R for a match, independent of the case of GIVEN_WORD

    • Finally, if GIVEN_WORD represents an complete expression, containing special regex characters, I advice you to use the following one : ^\h*(?-i)\QGIVEN_WORD\E(?-s).*\R. Indeed, between the 2 syntaxes \Q and \E, you may insert any character or symbol, without no restriction !!

    Refer to :

    For instance, if you want to delete the entire lines, beginning with the string +++ | Test | +++, whatever the case of the word test, and possibly preceded, in the same line, by some blank characters, use the regex, below :

    ^\h*(?i)\Q+++ | Test | +++\E(?-s).*\R

    Remark :

    If you don’t use the quoted form \Q.......\E, the regex must be rewritten as below, which each special regex character escaped with an \ character !

    ^\h*(?i)\+\+\+ \| Test \| \+\+\+(?-s).*\R

    Best Regards,


    P. S. :

    Claudia, many Thanks for your Xmas Greetings ! Sorry for not replying, in time :-< I just forgot to consult my e-mail box ;-))

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