thanks for all ur suggestions.
Can u please give me the sample code for ur solution.

Thanks

>On 11/14/06, Peter Boughton <[EMAIL PROTECTED]> wrote:
>> Ah, I knew I was missing something.
>> ....but... you can prevent that with a negative lookahead. :)
>> Something along the lines of.
>> <Emphasis type="([a-z])[^"]+">([^<]|<(?!Emphasis))+?</Emphasis>
>> So it only matches Emphasis tags that do not themselves contain the
>> <Emphasis string.
>> That should work, right? (inside an appropriate cfloop)
>
>In a pure world, then maybe.  But in the world created by this
>example, the tags *do* contain other tags of the same type.  This
>would match a tag if, and only if, it did *not*.  The tags wouldn't
>get replaced if other tags of the same type were nested inside.
>
>The regex you applied would, in effect, find the innermost Emphasis
>tag of a particular group, but there may be many levels of nest and
>even many adjacent tags.  Everything isn't quite that neat, I don't
>think.  :-)
>
>Given:
>
><Book>
><Emphasis type="Bold">
>sample text
> <Emphasis type="Italic">
>  sample Text
>       <Emphasis type="underline">
>           sample text
>       </Emphasis>
> </Emphasis>
>sample text
></Emphasis>
><Emphasis type="Bold">
>sample text
> <Emphasis type="Italic">
>  sample Text
>       <Emphasis type="underline">
>           sample text
>       </Emphasis>
> </Emphasis>
>sample text
></Emphasis>
></Book>
>
>Your regex would find the innermost tag of the first group, but
>without the help of additional code it wouldn't be able to adjust to
>the next, non-nested group nor would it be able to jump to the
>adjacent group.
>
>I think you're right - it could be done, but it doesn't seem as
>intuitive to my brain.  To me it's easier to find a particular tag
>(using /<Emphasis type="[a-z]+">/ or whatever regex) and from the end
>point of that tag look for /<\/?Emphasis>/.  If you encounter another
>open Emphasis tag, increment the counter, if you encounter a close
>Emphasis tag and the counter is "0" then you've found the end tag
>you're looking for, else decrement the counter.
>
>Like anything else, I guess:  there's more than one way to do it.  I
>just can't quite wrap my brain around the way you're proposing.  :-)

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