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(_dot_)Ross(_at_)iie(_dot_)qld(_dot_)gov(_dot_)au
-----Original Message-----
From: s-oualid(_at_)artefrance(_dot_)fr
[mailto:s-oualid(_at_)artefrance(_dot_)fr]
Sent: Friday, 21 February 2003 8:08 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] 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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