StringRegExpReplace, Several patterns

[Uten]

Now that topic tile was probably not self describing. This is what I want.

$r = StringRegExpReplace('This is a test "regarding" "back references" in a string', '"regarding"|"back references"', "$arg[\??]")
if $r = "This is a test $arg[1] $arg[2] in a string" Then 
    ; OK It works
Else
    ; hrmf, is it impossible?
EndIF

Replace \?? so it will return the back reference number (match number?) in the regexp pattern.

I think I have seen this in a sed script, and it might be possibly in perl. But does anyone know how to do it in pcre/Autoit?

[Xenobiologist]

Hi,

do you mean this?

$r = StringRegExpReplace('This is a test "regarding" "back references" in a string', '"regarding"|"back references"', "$arg[$0$1]")
ConsoleWrite($r & @CRLF)

So long,

Mega

[Uten]

No not exactly. I want to get the number of the back reference (hmm, at least I think thats what I want.. ). The test should pass.

if $r = "This is a test $arg[1] $arg[2] in a string" Then

[Xenobiologist]

No not exactly. I want to get the number of the back reference (hmm, at least I think thats what I want.. ). The test should pass.

if $r = "This is a test $arg[1] $arg[2] in a string" Then

HI,

I do not get the sense in that. If you know you replace two strings then why not just write it the way it is?

So long,

Mega

[Uten]

I want to parse and replace with variable references. I tried to provide the simplest possible test case. But I understand it might not have been to good or make sense to others.

Thanks for your input anyway. Actually I did not remember that $1..$9 (I'm using \1--\9)was the back reference in Autoit (and it is not documented as fare as I can see now) but I also think that $0$1 returns an unexpected result?

Why don't this work?

$r = StringRegExpReplace('This is a test "regarding" "back references" in a string', '"regarding"|"back references"', "$arg[$1]")
ConsoleWrite($r & @CRLF)

[Sokko]

Actually I did not remember that $1..$9 (I'm using \1--\9)was the back reference in Autoit (and it is not documented as fare as I can see now)

It's 0 to 9, not 1 to 9. Both $ and \ are valid for back-references in StringRegExpReplace and they are both documented in the parameters section. In this case you have only one group in the pattern, so 0 is the only back-reference that makes sense.

(EDIT: Forum eats backslash-zero)

Edited June 12, 2007 by Sokko

[MrCreatoR]

I think what you mean, is that will be multireplacment for all founded strings...

Try this function:

expandcollapse popup

$String = 'This is a test "regarding" "back references" in a string'
Dim $Pattern[2] = ['"regarding"', '"back references"']
Dim $Replace[2] = ['$arg[1]', '$arg[2]']

$String = _StringMultiReplace($String, $Pattern, $Replace)

MsgBox (64, "Results", $String  & @LF & "Replacment number: " & @extended)

Func _StringMultiReplace($String, $Pattern, $Replace)
    If Not IsArray($Pattern) Or Not IsArray($Replace) Then Return SetError(1, 0, $String)
    Local $Extended = 0
    ReDim $Replace[UBound($Pattern)]
    For $i = 0 To UBound($Pattern)-1
        $String = StringReplace($String, $Pattern[$i], $Replace[$i])
        $Extended += @extended
    Next
    Return SetError(0, $Extended, $String)
EndFuncoÝ÷ Øò¢çhpj{]y©pØZ®¶²²Úò¶¬jëh×6$String = 'This is a test "regarding" "back references" in a string'
$Patern = '"regarding"|"back references"'
$Replace = '$arg[1]|$arg[2]'

$String = _StringMultiReplace($String, $Patern, $Replace)

MsgBox (64, "Results", $String & @LF & "Replacment number: " & @extended)

Func _StringMultiReplace($String, $Patern, $Replace)
    If StringInStr($Patern, '|') And StringInStr($Replace, '|') Then
        Local $PaternArr = StringSplit($Patern, '|')
        Local $ReplaceArr = StringSplit($Replace, '|')
        Local $Ret = $String, $Extended = 0
        Local $Ubound = UBound($ReplaceArr)
        If UBound($PaternArr) <= UBound($ReplaceArr) Then $Ubound = UBound($PaternArr)
        For $i = 1 To $Ubound-1
            $Ret = StringRegExpReplace($Ret, $PaternArr[$i], $ReplaceArr[$i])
            $Extended += @extended
        Next
        Return SetExtended($Extended, $Ret)
    Else
        Return StringRegExpReplace($String, $Patern, $Replace)
    EndIf
EndFunc