>> <x>
>> <y Position='Top'/>
>> <y/>
>> </x>

"Martin Honnen" <mahotrash@yahoo.de> wrote in message
news:eGhsvjfQGHA.5400@TK2MSFTNGP09.phx.gbl...
>
>
> Martin wrote:
>
>> I have some data like:
>> <x>
>> <y Position='Top'/>
>> <y/>
>> <x/>
>
> Where is the closing </x> then? It is not clear whether that first x
> element has two y child elements as well as an x child element.
>
>> and a template like:
>> <xsl:template match="/x/y[@Position='Top']">
>> <li>
>> <xsl:if test="position()=last()">
>> <xsl:attribute name="class">last</xsl:attribute>
>> </xsl:if>
>> </li>
>> </xsl:template>
>>
>> I want to catch the last y with a position of Top
>> unfortunately position and last work on a list of y elements, not a list
>> of y[@Position='Top'] elements.
>
> The template does not change the type and size of the current node list.
> You simply need to use xsl:apply-templates to filter out nodes first e.g.
> <xsl:apply-templates select="y[@Position = 'Top']" />
>
> --
>
> Martin Honnen --- MVP XML
> http://JavaScript.FAQTs.com/

"Martin" <x@y.z> wrote in message
news:OMluiqfQGHA.2436@TK2MSFTNGP11.phx.gbl...
> Oops sorry, the data should read (1 x root, 2 y children)
>>> <x>
>>> <y Position='Top'/>
>>> <y/>
>>> </x>
>
> The rest of the problem description remains.
> Thanks
> Martin
>
> "Martin Honnen" <mahotrash@yahoo.de> wrote in message
> news:eGhsvjfQGHA.5400@TK2MSFTNGP09.phx.gbl...
>>
>>
>> Martin wrote:
>>
>>> I have some data like:
>>> <x>
>>> <y Position='Top'/>
>>> <y/>
>>> <x/>
>>
>> Where is the closing </x> then? It is not clear whether that first x
>> element has two y child elements as well as an x child element.
>>
>>> and a template like:
>>> <xsl:template match="/x/y[@Position='Top']">
>>> <li>
>>> <xsl:if test="position()=last()">
>>> <xsl:attribute name="class">last</xsl:attribute>
>>> </xsl:if>
>>> </li>
>>> </xsl:template>
>>>
>>> I want to catch the last y with a position of Top
>>> unfortunately position and last work on a list of y elements, not a list
>>> of y[@Position='Top'] elements.
>>
>> The template does not change the type and size of the current node list.
>> You simply need to use xsl:apply-templates to filter out nodes first e.g.
>> <xsl:apply-templates select="y[@Position = 'Top']" />
>>
>> --
>>
>> Martin Honnen --- MVP XML
>> http://JavaScript.FAQTs.com/
>
>

Martin wrote:

> I have some data like:
> <x>
> <y Position='Top'/>
> <y/>
> <x/>

> and a template like:
> <xsl:template match="/x/y[@Position='Top']">
> <li>
> <xsl:if test="position()=last()">
> <xsl:attribute name="class">last</xsl:attribute>
> </xsl:if>
> </li>
> </xsl:template>
>
> I want to catch the last y with a position of Top
> unfortunately position and last work on a list of y elements, not a list of
> y[@Position='Top'] elements.

Martin wrote:

> Oops sorry, the data should read (1 x root, 2 y children)

> The rest of the problem description remains.

>>
>>The template does not change the type and size of the current node list.
>>You simply need to use xsl:apply-templates to filter out nodes first e.g.
>> <xsl:apply-templates select="y[@Position = 'Top']" />

"Martin Honnen" <mahotrash@yahoo.de> wrote in message
news:ex01v5fQGHA.2704@TK2MSFTNGP15.phx.gbl...
>
>
> Martin wrote:
>
>> Oops sorry, the data should read (1 x root, 2 y children)
>
>> The rest of the problem description remains.
>
> Well I suggested a change in my earlier post
>
>
>>>
>>>The template does not change the type and size of the current node list.
>>>You simply need to use xsl:apply-templates to filter out nodes first e.g.
>>> <xsl:apply-templates select="y[@Position = 'Top']" />
>
>
>
> --
>
> Martin Honnen --- MVP XML
> http://JavaScript.FAQTs.com/