thank you umer...

now, what does the \1 mean in the 3rd parameter?

thanks.
tony


On Mon, 13 Dec 2004 19:02:10 -0500, Umer Farooq <[EMAIL PROTECTED]> wrote:
> reReplace(string,'(#chr(10)##chr(13)#)|(#chr(10)#|#chr(13)#|#chr(32)#)','\1','all'
> 
> Tony Weeg wrote:
> > so this reads:
> >
> > reReplace(string,'#chr(10)#|#chr(13)#|#chr(32)#','','all'
> >
> > replace all 10's 13's and 32's regardless of anything
> >
> > is there a way to say, dont remove where you find a (couplet)
> >
> > #chr(10)##chr(13)#
> >
> > just where you find either that arent together?
> >
> 
> -- 
> Umer Farooq
> Octadyne Systems
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