Jaime,
At 02:01 PM 1/13/2004, you wrote:
Thanks Wendell for your complete explanation. While reading your help, I
tested another alternative:
ROW[. = key('relacion_x_cobertura', REL_ID)[1]]
and it worked too. The actual XML structure has 35 ROW's, the first 7
ROW's has REL_ID = 1, the following 7 ROW's has REL_ID = 2, and so on
(this was a particular case where exist 7 ROW's for each REL_ID).
I think that the new way above worked because all of the ROW's are
different (besides REL_ID, the ROW has other fields), and since
key('relacion_x_cobertura', REL_ID)[1] returns only 1 node where the key
is equal to REL_ID, by comparing . with the key value, returns only 1 ROW,
the ROW that allows me to group by REL_ID=1, REL_ID=2, until REL_ID = 5.
I think that case is equivalent to generate-id method since all ROW's has
at least 1 field different, making the whole ROW different. Is that true?
Yes, if the *value* of each ROW is unique, it can likewise be used to
deduplicate. Note however this is a big IF, and isn't a reliable assumption
to make.
To understand this you need to know that $node1 = $node2 is true if the
(string) value of $node1 is the same as the string value of $node2 -- and
likewise, $nodeset1 = $nodeset2 is true if the value of *any* of the nodes
in $nodeset1 is the same as the value of any of the nodes in $nodeset2.
This is quite a loose and permissive way of construing node-set equality,
which is actually very useful sometimes. But it means that the "=" equality
test is not useful for testing node identity unless every node concerned
has a unique value -- which is rarely the case.
Cheers,
Wendell
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Wendell Piez
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Mulberry Technologies, Inc. http://www.mulberrytech.com
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Suite 207 Phone: 301/315-9631
Rockville, MD 20850 Fax: 301/315-8285
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