its a backreference.. to (#chr(10)##chr(13)#)

following page will explain better then I can.
http://www.regular-expressions.info/brackets.html


Tony Weeg wrote:
> 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
>>[EMAIL PROTECTED]
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>>
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> 
> 
> 

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