Re: [xsl] Find the Position Index of Elements‏‏

[David Carlisle <davidc@xxxxxxxxx>]

On 13/06/2010 20:43, Alice Wei wrote:
> Hi,
>
> I have an XML snippet as in the following:
>
> <music_songs>
>    <song>
>      <title>(I Just) Died In Your Arms</title>
>      <category>Rock</category>
>      <album>80 Popular Hits</album>
>      <artist>Cutting Crew</artist>
>      <date added="03-24-2009"/>
>    </song>
>    </music_songs>
>
> This is one of the songs out of categories I have in my xml file, and
> currently I use XPath expression as in the following:
> /music_songs/song[category='Rock' as an example to find all the songs
> in the Rock category.

whenever you use /a/b/c/[something='zzz']

you should think of doing

key('x','zzz')

instead, it can be surprising the speedup you can get
> I need to also be able to pull the position index of the song elements
> that I pull from the above expression, and use that in VB.NET for
> process. I tried using count(), position(), but they seem to be able
> to detect, but they cannot give me a list like

you don;t say if you are using xslt 1 or 2.

xslt 1 doesn't have a datatype that can store a sequence of integers.
You can get the numbers as for example

<xsl:for-each select="/music_songs/song[category='Rock']">

and then use

<xsl:number/>

or <xsl:value-of 
select="count(preceding-sibling::song[category='Rock'])+1)"/>

Ib xslt2 you can store a list of integers

/music_songs/song[category='Rock']/count(preceding-sibling::song[category='Rock'])+1)

for example would work (although is paerhaps rather inefficient)

David