xsl:number newb question

[Guiding5]

Hi group!

Basic question about counting specific nodes in the range, here the
example:

XSL:

<xsl:stylesheet version="1.0">
<xsl:template match="/">

Red cars:
<xsl:for-each select="car">
<xsl:if test="color = 'red'">
<xsl:number select="position()"/>.
<xsl:value-of select="manufacturer"/> ( <xsl:value-of
select="model"/> )
</xsl:if>
</xsl:for-each>

Blue cars:
<xsl:for-each select="car">
<xsl:if test="color = 'blue'">
<xsl:number select="position()"/>.
<xsl:value-of select="manufacturer"/> ( <xsl:value-of
select="model"/> )
</xsl:if>
</xsl:for-each>

</xsl:template>
</xsl:stylesheet>

I want to get XSLT result that will look like:
Red cars:
1. Ford (67)
2. AM (78)

Blue cars:
1. Fiat (92)
2. Toyota (06)
3. Ford (05)

But instead of it I get numbering of actualy nodes positions in
original XML.

[steve_marjoribanks]

I don't quite get what you're asking here, could you elaborate? What
number do you want if its not the node number?

[Martin Honnen]

Guiding5 wrote:

<xsl:for-each select="car">
<xsl:if test="color = 'red'">
<xsl:for-each select="car[color = 'red']">
I want to get XSLT result that will look like:
Red cars:
1. Ford (67)
2. AM (78)

should do that.

--

Martin Honnen
http://JavaScript.FAQTs.com/

[Guiding5]

Well, with given stylesheet I will get something like this (instead of
wanted):

1. Ford (67)
4. AM (78)

Blue cars:
2. Fiat (92)
3. Toyota (06)
5. Ford (05)

Counting will be according the real positions of elements in input XML.
The question is, how do I count elements in resulting XSLT ? How do I
get counting related to if condition ?

[Guiding5]

This actually did the job!

Thanks alot, it works, even if I still dont understand why :)

[Martin Honnen]

Guiding5 wrote:

it works, even if I still dont understand why :)

position() gives the position in the current node list, you had all car
elements in that list while my approach made sure the current node list
had only the red (or the blue) cars when position() is evaluated.
--

Martin Honnen
http://JavaScript.FAQTs.com/