-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of
Lorenzo De Tomasi
Sent: 17 February 2003 23:17
To: XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] problem with xsl:sort
I have a problem with xsl:sort
I have written this code
<xsl:for-each select=".">
<xsl:sort select="@type" data-type="text" order="ascending"/>
<xsl:for-each select="."> processes a node-set that always contains
exactly one node. So there isn't very much point in sorting it.
Michael Kay
Software AG
home: Michael(_dot_)H(_dot_)Kay(_at_)ntlworld(_dot_)com
work: Michael(_dot_)Kay(_at_)softwareag(_dot_)com
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XML
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="contact_temp.xsl"?>
<xCVP xmlns="" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:a="urn:oasis:names:tc:ciq:xsdschema:xAL:2.0"
xmlns:n="urn:oasis:names:tc:ciq:xsdschema:xNL:2.0"
xsi:schemaLocation="">
<PersonDetails lang="eng">
<Languages>
<Language type="other language" speak="good"
read="excellent" write="good" code="eng">English</Language>
<Language type="mother tongue" speak="excellent"
read="excellent" write="excellent" code="ita">Italiano</Language>
<Language type="other language" speak="basic"
read="basic" write="basic" code="fr">Fran?ßais</Language>
</Languages>
</PersonDetails>
</xCVP>
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XSL
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="xCVP/PersonDetails">
<html>
<body lang="eng">
<section>
<h>languages</h>
<p>
<xsl:apply-templates select="Languages/Language"/>
</p>
</section>
</body>
</html>
</xsl:template>
<xsl:template match="PersonInfo">
<xsl:apply-templates/>
</xsl:template>
<xsl:template match="Languages">
<xsl:apply-templates/>
</xsl:template>
<xsl:template match="Language">
<xsl:for-each select=".">
<xsl:sort select="@type" data-type="text" order="ascending"/>
<xsl:choose>
<xsl:when test="@type='mother tongue'">
<l><span class="label"><xsl:value-of
select="@type"/></span><xsl:text> </xsl:text><xsl:value-of
select="."/></l>
</xsl:when>
<xsl:otherwise>
<l><span class="label"><xsl:value-of
select="@type"/></span><xsl:text> </xsl:text><xsl:value-of
select="."/><xsl:text> </xsl:text><span
class="label">speak</span><xsl:text>
</xsl:text><xsl:value-of select="@speak"/><xsl:text>
</xsl:text><span class="label">read</span><xsl:text>
</xsl:text><xsl:value-of select="@read"/><xsl:text>
</xsl:text><span class="label">write</span><xsl:text>
</xsl:text><xsl:value-of select="@write"/></l>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
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and tested it in Netscape 7 for mac and the result is
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RESULT
languages
other language English speak good read excellent write good
mother tongue Italiano other language Français speak basic
read basic write basic
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As you can seen the result is not what I want. The lines are
not ordered by type attribute, but by xml order. Have I done
something wrong?
With xslt can I order a list created with xsl:for-each
specifying a precise order? i.e. I want to make a list and I
want to display first the items with attribute type="special
offer" in ascending order and then all the other ones by
ascending order.
XML
<item type="special offer!">book</item>
<item>book</item>
...
RESULT
special offer! book $10
special offer! cd $5
special offer! doll $10
book $20
cd $10
doll $20
Thank you very much!
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Lorenzo De Tomasi, student of Information Architecture,
Interface Design and Visual Design via Bellaria 6, 21018
Sesto Calende (Varese), Italia
phone: +39 (0)331 924649
mobile: +39 329 3941065; +39 333 8979304
e-mail: lorenzo(_dot_)detomasi(_at_)libero(_dot_)it;
lorenzo(_dot_)detomasi(_at_)email(_dot_)it
website: http://biografica.tzone.it
ICQ uin: 11313132
Yahoo! Instant Messenger id: tummait
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