Visual Studio - Using RegExp with Submatches in VS-Replace-dialog

Author

Using RegExp with Submatches in VS-Replace-dialog

Alexander Mueller

2006-02-13, 7:01 pm

Hi @all

is it possible to use submatches in the search & replace dialog of visual studio
2005? I want to replace uniform patterns of VB-6-code that deal with different
variable-names and different literals but have fully identical same structure,
example:

old VB-Code:

If (Left(strName, Len("SAB3_4_CREDIT")) = "SAB3_4_CREDIT" Then

If (Left(strLoan, Len("RNT_7_CREDIT")) = "RNT_7_CREDIT" Then

i want to change it to

If strName.StartsWith ("SAB3_4_CREDIT") Then

If strLoan.StartsWith ("RNT_7_CREDIT") Then

I know RegExp from Jscript 5.6 (WSH) and a pattern to find
this Left-Len-construct and building submatches for the varaible
and the literal goes like this:

var r = /$Left\((\w+), Len\(("[^"]+")$\) = "[^"]+" Then/;

the replace-expression for JScript is:

$1.StartsWith($2) Then

VStudio seesm to have the same syntax for finding-patterns.
But it seems to have a different syntax to denote submatches
inside the replace-expression, if any.
Is there support for submatches in the replace-expression, at all?

Thanks for your help,
Alex

Alexander Mueller

2006-02-14, 3:59 am

Alexander Mueller schrieb:

> I know RegExp from Jscript 5.6 (WSH) and a pattern to find
> this Left-Len-construct and building submatches for the varaible
> and the literal goes like this:
>
> var r = /$Left\((\w+), Len\(("[^"]+")$\) = "[^"]+" Then/;
>
> the replace-expression for JScript is:
>
> $1.StartsWith($2) Then
>
> VStudio seesm to have the same syntax for finding-patterns.
> But it seems to have a different syntax to denote submatches
> inside the replace-expression, if any.
> Is there support for submatches in the replace-expression, at all?

Thanks to Herfried I found that VStudio-online help uses the term
'tagged expressions' for submatches, curly brackets for building them
and \n for referencing them in a replace-pattern.

So changing all occurences

If Left(<varname>, Len("<string-literal>")) = "<string-literal>" Then

to

<varname>.StartsWith("<string-literal>") Then

can be by achieved by setting

search to: Left${:a+}, Len\({"[^"]+"}$\) = "[^"]+" Then
replace to: \1.StartsWith(\2)

and clicking Replace All.

just in case anybody else wants to migrate his old code to NET.

Mfg,
Alex

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