Vista - Replace Nth character in a string

Hello,

is it possible to replace a character at a arbitrary position in a string ?

For example :
I want to replace the third and hundredth character of the variable myText
with an * :

So bascically something like this:
mid(myText, 3,1) = "*"
mid(myText, 100,1) = "*"

I had a look at the "replace" function... but this needs a character to
find.... but I have only a position that I need to replace.

This is what I have tried :

[01]Function cleanString (sString)
[02] '** replace each character having ASC-Value <31 with "*" - execpt 10
and 13 (linefeed)
[03] for i = 1 to len (sString)
[04] iCharToCheck = asc ( mid (sString,i,1 ) )
[05] if iCharToCheck < 31 and ( iCharToCheck <> 10 and iCharToCheck
<> 13) then
[06] '** i = position to replace :
[07] sString = left(sString, i-1) & "*" & mid(sString, i+1,999)
'** concat everthing left and right from the occurance - place "*"
in-between
[08] wscript.echo "fixed " & mid (sString,i,1) & "@" & i
[09]
[10] end if
[11]
[12] next
[13]
[14] cleanString = sString
[15]
[16] end function

But this replaces the first occurence only - and then I get :

"invalid procedure call or argument: 'asc' " in line 2


thank you

small mistake in my posting:

"invalid procedure call or argument: 'asc' " in line 4


"Heinz" <Spacewalker4711(noSpam)@hotmail.com> schrieb im Newsbeitrag
news:uvY6%23tG4JHA.5048@xxxxxx

"Heinz" <Spacewalker4711(noSpam)@hotmail.com> wrote in message
news:uvY6%23tG4JHA.5048@xxxxxxIf you want to replace the 3rd and 100th characters with "*", then the Mid
function seems best. For example:

If (Len(strOld) > 100) Then
strNew = Left(strOld, 2) & "*" & Mid(strOld, 4, 96) & "*" & Mid(strOld,
101)
End If

If instead you want to replace all characters where the ASCII value is less
than 31 (except 10 and 13) with "*", then you need to look at every
character in the string. I would use:

For k = 1 To Len(strOld)
If (Asc(Mid(strOld, k, 1)) < 31) Then
If (Asc(Mid(strOld, k, 1)) <> 10) And (Asc(Mid(strOld, k, 1)) < >
13) Then
strOld = Left(strOld, k - 1) & "*" & Mid(strOld, k + 1)
End If
End If
Next

Your code looks similar, but I didn't get your error when I tested. I don't
know why, but I never use the extra spaces in the function calls, and I used
the default for the third parameter of the Mid function (which is the rest
of the string) when I don't know how long the rest of the string is.

--
Richard Mueller
MVP Directory Services
Hilltop Lab - http://www.rlmueller.net
--

thank you.. I've cut & paste your code and it works...

"Richard Mueller [MVP]" <rlmueller-nospam@xxxxxx> schrieb im
Newsbeitrag news:O0A7qOH4JHA.1420@xxxxxx