Re: Sorting a sequence and selecting the first element

You could also use the union operator, or its short form |, as in
(foo | *[descendant::foo | descendant::bar])[1]

In contrast to the sequence operator (the comma), it will eliminate
duplicates and return its results in document order.

Gerrit

On 08.05.2015 13:57, Jorge . chocolate.camera@xxxxxxxxx wrote:
> I answer both questions to myself:
> 
>     <xsl:sequence="*[local-name() eq 'foo' or descendant::foo or
> descendant::bar][1]"/>
> 
> (unless I got something wrong again).
> 
> Please, excuse me for the noise.
> 
> On Fri, May 8, 2015 at 1:50 PM, Jorge . chocolate.camera@xxxxxxxxx
> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>> I now see that, on top of that, I am adding duplicate copies of nodes
>> that match both conditions to the sequence (-_-;). I would also
>> appreciate a hint on how to make that selection without dupes.
>>
>> On Fri, May 8, 2015 at 1:44 PM, Jorge . chocolate.camera@xxxxxxxxx
>> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>>> I'd like to select the first child of the current node that either:
>>>
>>> 1.    is a "foo" element
>>> 2.    has "foo" or "bar" descendant(s)
>>>
>>> The following sequence returns me all nodes matching either or both of
>>> those conditions:
>>>
>>>     <xsl:sequence select="foo, *[descendant::foo, descendant::bar]"/>
>>>
>>> but not necessarily ordered according to their position as children of
>>> the current node. As a result, the first element in that sequence,
>>> selected like this:
>>>
>>>     <xsl:sequence select="(foo, *[descendant::foo, descendant::bar])[1]"/>
>>>
>>> is not necessarily the first matching child of the current node.
>>>
>>> How do I sort elements in that sequence according to their position as
>>> children of the current context, so that the first element in that
>>> sequence is actually the first matching child in the current node?
>>>
>>
> 
> 

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