rereplace(string ,"(\s)[\1\s]+","\1""all") Might that do the trick? Thats gonna replace any space character that is followed by another space character (or more) with just the first space char found...or atleast tthats what I was goiong for not positive on the syntax and nothing to check with ATM...
Adam H On Mon, 13 Dec 2004 16:44:07 -0500, Tony Weeg <[EMAIL PROTECTED]> 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? > > -- > tony > > Tony Weeg > > macromedia certified coldfusion mx developer > email: tonyweeg [at] gmail [dot] com > blog: http://www.revolutionwebdesign.com/blog/ > cool tool: http://www.antiwrap.com > > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~| Special thanks to the CF Community Suite Silver Sponsor - CFDynamics http://www.cfdynamics.com Message: http://www.houseoffusion.com/lists.cfm/link=i:4:187468 Archives: http://www.houseoffusion.com/cf_lists/threads.cfm/4 Subscription: http://www.houseoffusion.com/lists.cfm/link=s:4 Unsubscribe: http://www.houseoffusion.com/cf_lists/unsubscribe.cfm?user=89.70.4 Donations & Support: http://www.houseoffusion.com/tiny.cfm/54
