Re: [xsl] how to get position of node in node-set

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Erwin,

At 12:32 PM 9/11/2003, David wrote:
> Node sets are sets and so don't have an intrinsic order.
> So nodes don't really have a position in a set (you can't for example
> sort the nodes and save the sorted list in a set)

Of course you can try, but then in XSLT you have a result-tree-fragment, 
not a node set, which is where my answer (of a few minutes ago) starts. 
(Sorry if this assumption on my part was incorrect.)

In XSLT 1.0, whether a variable refers to a node set or to an RTF, depends 
on how it's created. For example, if (using a key-based technique to 
de-duplicate) you declare your variable like this:

<xsl:variable name="legend"
   select="//entry[count(.|key('entries-by-text',@text)[1])=1]"/>

you'll have a true node set, in which case all David's admonitions apply. 
They are not in order within $legend, and if you process them using 
for-each or apply-templates, they'll be processed in the order they appear 
in the source document.

If, however, in order to sort the nodes, you say

<xsl:variable name="legend">
  <xsl:apply-templates select="//entry" mode="de-duplicate">
    <xsl:sort select="@text"/>
  </xsl:apply-templates>
</xsl:variable>

you'll get your order but you won't have a node set, rather an RTF.

To convert an RTF to a node-set requires the extension function in XSLT 
1.0. Then you can operate on it again. The only things you can do with an 
RTF are copy it to the result (as many times as you like :-), or turn it 
into a string and do stuff with that.

(David continues)
> You can ask what is the position if the nodes in the set were sorted
> into document order, although actually it's not so easy.
> If the nodes are siblings and you are on <entry text="ccc"/>
> then you want
> count(preceding-sibling::entry)+1
> however you only want to count nodes that are in $legend
> so that would be
> count(preceding-sibling::entry[@text=$legend])+1

Unfortunately this counts the preceding siblings whose @text attribute has 
a string value equal to the value of some node in $legend ... maybe not so 
useful.

> if your text attribute uniquely determines the node,

Right (but Erwin said it wasn't: bummer).

>  or something like
> count(preceding-sibling::entry[count(.|$legend)=count($legend)]+1
> otherwise.

That's very sneaky, and might just work. *If* your nodes are siblings (I'm 
betting they're not).

> They are all pure xpath solutions, or you could use xsl, as in
>
> <xsl:for-each select="$legend">
> <xsl:if test="@text='ccc'"><xsl:value-of select="position()"/></xsl:if>

Which suggests a way to avoid the node-set() extension ... assuming $legend 
is a node set (which has no intrinsic order):

<xsl:template match="entry"/>
  <xsl:variable name="text" select="@text"/>
  <xsl:for-each select="$legend">
    <xsl:sort select="@text"/>
    <xsl:if test="@text=$text">
      <xsl:value-of select="position()"/>
    </xsl:if>
  </xsl:for-each>
</xsl:template>

Who said it's not possible in XSLT 1.0? (Thanks, David.)

Will the language be easier or harder when RTFs disappear and become node 
sequences? Who's to say?

Cheers,
Wendell


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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
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