Help with regular express

[lsakizada]

I am not expert with regular express, please help.

How do I get the path part from a string that contain path and other haracters such as by using the stringRegExp() method?

C:\AutoMDP\AE27.au3 - 1 error(s), 0 warning(s)

.

I need this part only:

C:\AutoMDP\AE27.au3

Note: The location of the au3 file may have different location.

and file name may differ as well.

Edited February 23, 2008 by lsakizada

[MrCreatoR]

Hi,

Try this:

$String = "C:\AutoMDP\AE27.au3 - 1 error(s), 0 warning(s)"

$Results = StringRegExpReplace($String, "(.*?) - \d .*$", "\1")

ConsoleWrite($Results)

[lsakizada]

Hi,

Try this:

$String = "C:\AutoMDP\AE27.au3 - 1 error(s), 0 warning(s)"

$Results = StringRegExpReplace($String, "(.*?) - \d .*$", "\1")

ConsoleWrite($Results)

MsCreatoR! You have shoot before I pushed the Post button!

Absolutely returns the expected string. I was not expected for better.

thanks my friend!

[lsakizada]

Oops! sorry. This is not enough for my needs, although i was sure at the beginning.

The problem is that I need to get the path from the string regardless the tailing chars after the path.

The solution you have gave me is dippend on the number that came after the path...

Please help

[rasim]

Mayb this?:

$String = "C:\AutoMDP\AE27.au3 - 1 error(s), 0 warning(s)"

$Results = StringRegExp($String, "^.*\.au3", 1)

MsgBox(0, "", $Results[0])

[lsakizada]

Mayb this?:

$String = "C:\AutoMDP\AE27.au3 - 1 error(s), 0 warning(s)"

$Results = StringRegExp($String, "^.*\.au3", 1)

MsgBox(0, "", $Results[0])

Thanks Rasim! I have tested it with my application and it is looks good so far.

Have you know free application that whould do the regexp easier?

Thank you very much.

[rasim]

Have a book Mastering Regular Expressions J.E.F. Friedl

[lsakizada]

Have a book Mastering Regular Expressions J.E.F. Friedl

THANKS AGAIN!