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Re: [xsl] calculating relative position in document order

2008-10-16 14:16:15
Hi Tom,

(Earlier today Michael Kay answered a message that involved
recommending this library:  http://www.functx.com/functx/ ... so I had
a quick look at functx and it is fresh in my mind.)

These functions are not exactly like your fantasy pseudo code... but I
think there some useful functions that may have help solve your
problem....

For example:
http://www.xsltfunctions.com/xsl/functx_sequence-node-equal.html
http://www.xsltfunctions.com/xsl/functx_sequence-node-equal-any-order.html

If you peruse the functions, I think you'll find many other
interesting/related functions that you may be able to leverage.

(Note: I found the download link difficult to find on the page...
Hint: It's at the top of the page.)

Darcy
On Thu, Oct 16, 2008 at 12:53 PM, tom s <tshmit(_at_)yahoo(_dot_)com> wrote:
Hi,

I'd like to be able to determine which element is "closer", in terms of 
document order, to a given element. eg, given two structures containing 
siblings:

       <a>
       <b>
       <c>
and
       <b>
       <a>
       <c>

I'd like to detect that the second structure is "out of order". In my fantasy 
pseudo code, I might express the constraint as:

//c[positionEX(preceding::a[1]) > positionEX(preceding::b[1])]

Where my fantasy positionEX() function returns the absolute position of the 
argument in the document.

Am I missing something, or is implementing this operation in real xpath/xslt 
much more difficult than it seems it should be?

Thanks in advance for any help...

--T




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