Old StringRegExp

[ParoXsitiC]

I am renewing an older library so that it works with the current.

$iNumChrsToRead = StringRegExp($sParamsInStr, "(\d*?)(\#)(?:\$\[)", 1)
$sParamsInStr = StringTrimLeft($sParamsInStr, $iNumChrsToRead[1])
While StringLen($sParamsInStr) < $iNumChrsToRead[0]

Where $sParamsInStr is equal to "49$[0]$"11111111"$[0]$<nil>$[0]$"2222222"$[0]$<nil>"

I am pretty sure he wants $iNumChrsToRead[1] to be 2

and $iNumChrsToRead[0] to be 49.

What is throwing me off is the use of (\#) .. it seems like at one time would return the length of the first catch. Anyone familar with (\#)

Current AuotIT throws an error because it doesn't find a match to his regex

[PsaltyDS]

All this "(\#)" means to PCRE is to capture an escaped pound sign as a group.

Are you sure you meant to apply those double quotes that way?

Modify this demo to show the exact value intended for $sParamsInStr, and what you meant to receive in $aNumChrsToRead:

#include <Array.au3>

$sParamsInStr = '"49$[0]$"11111111"$[0]$<nil>$[0]$"2222222"$[0]$<nil>"'
$aNumChrsToRead = StringRegExp($sParamsInStr, "(\d*?)(\#)(?:\$\[)", 1)
If @error Then
    ConsoleWrite("Debug:  @error = " & @error & "; @extended = " & @extended & @LF)
Else
    _ArrayDisplay($aNumChrsToRead, "$aNumChrsToRead")
EndIf

As is, it gets no matches because there are no literal "#" characters in the string.

[ParoXsitiC]

I had asked about the meaning \# back in 2007. AutoIT forums don't keep that old of history it seems. The double quotes we letting you know it was a string. It doesnt have "" in it. Either way, I just replaced the idea of \# and used StringLen instead.

[GEOSoft]

You wouldn't by any chance be thinking of \x## where ## can be 2 hexadecimal digits would you, or even the \### where ### can be up to 3 octal digits?

Literal double quotes can be handles with \x22 for example. As was already ponted out \# just means the literal # symbol.

Edit. For future reference, remember that SRE arrays are 0 based.

$iNumChrsToRead = StringRegExp($sParamsInStr, "(\d*?)(\#)(?:\$\[)", 1);; Return the first match as an array.

will return an array where $iNumChrsToRead[0] will be all digits and $iNumChrsToRead[1] will be the # sign. The expression will not match at a $ symbol followed by a [ symbol

So The type of string it would be looking for would be

aabbaa123456#[

Where element 0 is 123456

and element 1 is #

and the # symbol must be followed by $[ in the original string

Edited July 12, 2010 by GEOSoft