You are changing the input string after every replace. I assume that is the reason for the strange results. Use a temporary string (stroutput). When everything is replaced you set strinput = stroutput.

Boris

On 1 mar 2006, at 05:27, vvor wrote:

this helps awesomely!
except - how do i replace in place the original strinput? do i keep a position counter? do you have a simple example of that, and then i promise i won't bother anybody.
i tried it with a counter, and i'm failing:

strinput = rm.replace(replacementpattern, positioncounter)

and everything gets offset in weird ways.

:(

-vv

On Feb 26, 2006, at 2:52 AM, dda wrote:

Do it in a loop:

Dim rm As RegexMatch
rm=r.search(strinput)
while rm<>Nil
  <do whatever you have to do>
rm=r.search(strinput,rm.SubExpressionStartB(0)+LenB(rm.SubExpressionSt ring(0)))
wend

HTH

--
dda
libcurl4RB, [S]FTP transfers made easy
http://sungnyemun.org/?q=node/8

RBDeveloper Columnist, "Beyond the Limits"
http://rbdeveloper.com

On 2/26/06, vvor <[EMAIL PROTECTED]> wrote:
thank you, this works. but what if i need access to $1, $2, etc.,
outside of the r.replace? i'd like to process the replacement pattern
a bit based on conditions prior to the replacement. this is done with
regexmatch, yes? but that only matches once?

iow, i want to remember if i match a certain thing so i can alert
other code so that i can influence my replacement patterns a bit.

sorry if i'm dense.

-vv
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