RE: xsl:sort and axis navigation

Within a template it takes the original (XML) document ordering.  You can use a for-each to access a sorted order if desired.

HTH,

Ken Ross
Ph: +61 7 32359370
Mob: +61 (0)419 772299
Email: Ken.Ross@xxxxxxxxxxxxxx


-----Original Message-----
From: s-oualid@xxxxxxxxxxxxx [mailto:s-oualid@xxxxxxxxxxxxx]
Sent: Friday, 21 February 2003 8:08 AM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject:  xsl:sort and axis navigation


Hello !

I'm new here and I have a little question about xsl sorting and axis navigation with XPath, I checked the faq and did'nt find the answer (even if I guess that's a common question...).

I do a test in a <xsl:template match="FICHE"> tag, when I call this 
template, I do a sort :

<xsl:apply-templates select="FICHE">
<xsl:sort select="SOUSTHEME" />
</xsl:apply-templates>

Then in the template, when I this test :

<xsl:if test="SOUSTHEME != following-sibling :: SOUSTHEME[1]">
<xsl:attribute name="break-after">page</xsl:attribute>
</xsl:if>

Does the test take the sorted data or the original XML datas (coming from a java classes) ?

In other words, should I do a first transformation to sort my data in 
order to be able to do this test correctly ?

I think I should because the result is not what I expected... I just need a confirmation, or maybe a way for me not to do 2 XSLT transformation.

Thanks !

Simon

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