RegExp Find & Replace 4.3 w/ 253 presets (2011-07-06) [MM2+]

[Owyn]

Is there any chance for a string to fix all the tags in one step?

Code: Select all

\s\[[^\]]*\]/.*?$

Ummm (not tested).
Into:Title
Find What:(^.*\))(.*$)
Replace With:$1
Both regexp checked.

This should keep the start of the title including the first parenthesized string.

Edit: This was a first thing in the morning brain fart. Guest's solution should be exactly what you are looking for.

[Guest]

This should keep the start of the title including the first parenthesized string.

Yeah, but what with [feat. Mr. V] in: Back & Forth (Eric Chase RMX) [feat. Mr. V] [Eric Chase RMX]/Eric Chase RMX?

I suppose he wants to get Back & Forth (Eric Chase RMX) [feat. Mr. V], not only Back & Forth (Eric Chase RMX).

[Owyn]

Yeah. I thought of that, but, It breaks down when the genre string is only a part of the artist string.

Sorry, but I don't understand what you meant.

e.g. Artist="2pac feat. somebody_else), etc.

[Owyn]

This should keep the start of the title including the first parenthesized string.

Yeah, but what with [feat. Mr. V] in: Back & Forth (Eric Chase RMX) [feat. Mr. V] [Eric Chase RMX]/Eric Chase RMX?

I suppose he wants to get Back & Forth (Eric Chase RMX) [feat. Mr. V], not only Back & Forth (Eric Chase RMX).

Yeah. Your regexp should strip the trailing, probably duplicate, text. Better fit.

[Guest]

e.g. Artist="2pac feat. somebody_else), etc.

Well, if he has genre= 2pac feat. somebody_else; Rap; Hip Hop, it will be removed as well. However, if he has artist=2pac feat. somebody_else and genre=2pac; Rap; Hip Hop then it is too complicated.

[Guest]

Yeah. Your regexp should strip the trailing, probably duplicate, text. Better fit.

Sorry, but I think that my regexp does exactly what he wants. Also, your regexp is not good with titles like "(I Can't Get No) Satisfaction" and similar where parenthesized text is not on the end of the string.

[Owyn]

Yeah. Your regexp should strip the trailing, probably duplicate, text. Better fit.

Sorry, but I think that my regexp does exactly what he wants. Also, your regexp is not good with titles like "(I Can't Get No) Satisfaction" and similar where parenthesized text is not on the end of the string.

Agreed.

[Kevinowpb]

e.g. Artist="2pac feat. somebody_else), etc.

Well, if he has genre= 2pac feat. somebody_else; Rap; Hip Hop, it will be removed as well. However, if he has artist=2pac feat. somebody_else and genre=2pac; Rap; Hip Hop then it is too complicated.

I have fixed most of my feat. artists to be 2pac; Dr Dre with your RegExp find feat., vs., &, and replace with ;

so.. I'll give your theory a shot..

Isnt working...

Here is what I have... what's wrong?

[Guest]

Here is what I have... what's wrong?

Regular expression 1 (below the Into dropdown) should be checked on.

[Guest]

so.. I'll give your theory a shot..

Anyway, it will not work if your artist name is different than the single item form genre. It doesn't matter if featuring artists are separated with "feat." or with ";". So, if you have Artist = "2pac; Dr Dre" and Genre = "2pac; Rap; Hip Hop" then it will not be removed with mentioned suggestion.

[Guest]

OK, if you have multi-item artists separated with ";", you should try with the next Replace with string:

Code: Select all

Mid(RegSub(";$&", ";\s?(" & RegSub(RegSub(oSongData.ArtistName, "(\W)", "\$1"), ";\s?", "|") & ")", ""), 2)

[Guest]

Sorry, the previous expression is not proper. Here is the correct one:

Code: Select all

LTrim(Mid(RegSub(";$&", ";\s?(" & RegSub(RegSub(oSongData.ArtistName, "([^;\w])", "\$1"), ";\s?", "|") & ")", ""), 2))

[SpritHansi]

Hi!
I want to replace all { & [ with (, and ] & } with )
Help....

[Guest]

I want to replace all { & [ with (, and ] & } with )

Find what: (\[|{)|(\]|})
Regexp 1: checked
Replace with: MapArray(Array("$1", "(", "$2", ")"))
VBScript: checked

[Owyn]

Replace with: MapArray(Array("$1", "(", "$2", ")"))

Nice. I would have wimped out with two separate operations.