Re: XPath to find duplicate elements

Just a thought on this.

If we can implement deep-equal($n1, $n2)  as calculating some
ValueIdentity (VI) for each of the nodes and returning

     VI1 eq VI2 ...

Then the VIs for each node can be calculated on the initial reading of
the XML document (even in streaming mode) with almost no additional
effort. If these VIs are made properties of the nodes, then we could
just compare the VIs of the nodes and avoid any repeated zip-like
comparison of the subtrees that are topped by these nodes. And this
also seems more declarative/functional.

Just my 2c.

Cheers,
Dimitre



On Thu, Apr 5, 2018 at 9:45 AM, Martin Honnen martin.honnen@xxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
> On 05.04.2018 18:12, Leo Studer leo.studer@xxxxxxxxxxx wrote:
>
>> I have an xml file where I need and XPath expression to find all elements
>> that are a copy of a previous element in the same file.
>>
>> I came up with the following:
>>
>>
>> *let* *$xml* := <a>
>> <b>
>> <c>1</c>
>> <d>1</d>
>> <d>2</d>
>> </b>
>> <b>
>> <c>2</c>
>> <d>1</d>
>> <d>3</d>
>> </b>
>> <b>
>> <c>3</c>
>> <d>2</d>
>> <d>3</d>
>> </b>
>> <b>
>> <c>3</c>
>> <d>2</d>
>> <d>3</d>
>> </b>
>> </a>
>> *return*
>>
>> *$xml*//*[*let**$node1*:=.*return**some**$node2**in**$xml*//*[.<<*$node1*]
>>
*satisfies*/deep-equal/(*$node1*,*$node2*)]/(ancestor-or-self::/node/()//conc
at/(/node-name/(.),1+/count/(preceding-sibling::*)),'&#10;')
>>
>> which gives
>>
>>   a1 b2 d2
>>   a1 b3 d2
>>   a1 b3 d3
>>   a1 b4
>>   a1 b4 c1
>>   a1 b4 d2
>>   a1 b4 d3
>>
>>
>> This is actually not to bad. However, I would like the correct position of
>> the copy. Instead of preceding-sibling::/* /in the count() function, I
would
>> like to write something like preceding-sibling::/element(node-name(.)).
>> /Unfortunately the processor does not allow that ;-(b&
>
>
> The code with $xml := <a>...</a> looks like XQuery to me, not like XSLT or
> XPath. That aside, what do you want pseudo syntax
>   preceding-sibling::element(node-name(.))
> to compute? If you want to select all elements and compare them to the
> node-name() of another node you can certainly use
>   preceding-sibling::*[node-name() = node-name($n)]
> where you simply would need to make sure you define $n previously with a
> "let", as you have done at another place in your code.
>
> Also XPath now has a "path" function which returns an XPath expression with
> positional predicates so perhaps you don't need to construct the
> name/position pairs on your own if you have access to the latest XPath
> version.
>



--
Cheers,
Dimitre Novatchev
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