problem with regexp

[j0kky]

Hi,

I don't understand why this expression is not lazy:

$path = "some\path\that\annoys"
ConsoleWrite(StringRegExpReplace($path, "\\.+?$", "", 0) & @CRLF)

I'm trying to delete just "annoys" but the regexp matches "paththatannoys" 

[guinness]

Because like humanity it's greedy. Use (?U).

[guinness]

Don't listen to the guy in post #2 (he's stupid!)...try this...

$path = "some\path\that\annoys"
ConsoleWrite(StringRegExpReplace($path, "\\[^\\]*$", "") & @CRLF)

Edited April 22, 2014 by guinness

[UEZ]

Here my version:

ConsoleWrite(StringRegExpReplace($path, "(.+)\\.*", "\1") & @CRLF)

@guinness: your regex will fail to remove last <word> when <word> is empty.

 

 

Br,

UEZ

Edited April 22, 2014 by UEZ

[guinness]

  On 4/22/2014 at 6:21 PM, UEZ said:

@guinness: your regex will fail to remove last <word> when <word> is empty.

Fixed it! Though in all honesty I was only providing what the OP suggested.

[mikell]

  On 4/22/2014 at 5:40 PM, j0kky said:

Hi,

I don't understand why this expression is not lazy:

 

Cruel people, providing solutions without answering the question... 

".+?$" : the regex engine finds the first antislash, then returns the shortest match up to the end (lazy or not lazy don't matter) because the dot in .+?  matches everything including the next antislashes

That's why guinNess solution [^]* is the good one (matching everything EXCEPT antislash)

Edit

(unpardonable typo)

Edited April 22, 2014 by mikell

[guinness]

  On 4/22/2014 at 8:50 PM, mikell said:

That's why guinNess solution [^]* is the good one (matching everything EXCEPT antislash)

Brilliant!

[j0kky]

Thank you at all! I've understood my mistake