Advanced Command-Line Replace Command In VBScript - command-line

Question

I'm writing a compiler for my won computer language. Now before the language can be compiled i actually need to replace all apostrophes (') with percents (%) via a command-line vbs program. But the apostrophes only need to be replaced if there is NOT a circumflex accent (^) in front of it. So for example, in this code:
color 0a
input twelve = 0a "hi! that^'s great! "
execute :testfornum 'twelve'
exit
:testfornum
if numeric('1) (
return
) ELSE (
print 0a "oops 'twelve' should be numeric"
)
return
the apostrophe at line 2 should not be replaced, but the ones at line 3, 6 and 9 should be.
can anyone help me?
this is what i have so far:
'syntax: (cscript) replace.vbs [filename] "StringToFind" "stringToReplace"
Option Explicit
Dim FileScriptingObject, file, strReplace, strReplacement, fileD, lastContainment, newContainment
file=Wscript.arguments(0)
strReplace=WScript.arguments(1)
strReplacement=WScript.arguments(2)
Set FileScriptingObject=CreateObject("Scripting.FileSystemObject")
if FileScriptingObject.FileExists(file) = false then
wscript.echo "File not found!"
wscript.Quit
end if
set fileD=fileScriptingobject.OpenTextFile(file,1)
lastContainment=fileD.ReadAll
newContainment=replace(lastContainment,strReplace,strReplacement,1,-1,0)
set fileD=fileScriptingobject.OpenTextFile(file,2)
fileD.Write newContainment
fileD.Close

Answer 1

As #Ansgar's solution fails for the special case of a leading ' (no non-^ before that), here is an approach that uses a replace function in a test script that makes further experiments easy:
Option Explicit
Function fpR(m, g1, g2, p, s)
If "" = g1 Then
fpR = "%"
Else
fpR = m
End If
End Function
Function qq(s)
qq = """" & s & """"
End Function
Dim rE : Set rE = New RegExp
rE.Global = True
rE.Pattern = "(\^)?(')"
Dim rA : Set rA = New RegExp
rA.Global = True
rA.Pattern = "([^^])'"
'rA.Pattern = "([^^])?'"
Dim s
For Each s In Split(" 'a^'b' a'b'^'c nix a^''b")
WScript.Echo qq(s), "==>", qq(rE.Replace(s, GetRef("fpR"))), "==>", qq(rA.Replace(s, "$1%"))
Next
output:
cscript 25221565.vbs
"" ==> "" ==> ""
"'a^'b'" ==> "%a^'b%" ==> "'a^'b%" <=== oops
"a'b'^'c" ==> "a%b%^'c" ==> "a%b%^'c"
"nix" ==> "nix" ==> "nix"
"a^''b" ==> "a^'%b" ==> "a^'%b"

Answer 2

You can't do this with a normal string replacement. A regular expression would work, though:
...
Set re = New RegExp
re.Pattern = "(^|[^^])'"
re.Global = True
newContainment = re.Replace(lastContainment, "$1%")
...