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Old July 2nd, 2012, 04:11 AM
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Default sorting within current-group()

Hi,

This is my first question (post) in this forum, I wish to sort the <cityNames> within the <record> element. Is there any way to sort?

Input:

<cities>
<cityName>Barodara</cityName>
<cityName>Ahemadabad</cityName>
<stateName>Gujrat</stateName>
<cityName>Varanasi</cityName>
<cityName>Allahabad</cityName>
<stateName>Uttar Pradesh</stateName>
<cityName>Hyderabad</cityName>
<cityName>Bangalore</cityName>
<stateName>Andhara Pradesh</stateName>
</cities>


XSL Code:

<xsl:template match="cities">
<xsl:copy-of select="rit:cit(./*)"/>
</xsl:template>
<xsl:function name="rit:cit" as="item()*">
<xsl:param name="ct" as="node()*"/>
<xsl:element name="root">
<xsl:for-each-group select="$ct" group-ending-with="stateName">
<xsl:element name="record">
<xsl:copy-of select="current-group()"/>
</xsl:element>
</xsl:for-each-group>
</xsl:element>
</xsl:function>

Current Output:

<root>
<record>
<cityName>Barodara</cityName>
<cityName>Ahemadabad</cityName>
<stateName>Gujrat</stateName>
</record>
<record>
<cityName>Varanasi</cityName>
<cityName>Allahabad</cityName>
<stateName>Uttar Pradesh</stateName>
</record>
<record>
<cityName>Hyderabad</cityName>
<cityName>Bangalore</cityName>
<stateName>Andhara Pradesh</stateName>
</record>
</root>


Expected Output:
<root>
<record>
<cityName>Ahemadabad</cityName>
<cityName>Barodara</cityName>
<stateName>Gujrat</stateName>
</record>
<record>
<cityName>Allahabad</cityName>
<cityName>Varanasi</cityName>
<stateName>Uttar Pradesh</stateName>
</record>
<record>
<cityName>Bangalore</cityName>
<cityName>Hyderabad</cityName>
<stateName>Andhara Pradesh</stateName>
</record>
</root>


thanks!
 
Old July 2nd, 2012, 04:20 AM
mhkay's Avatar
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Default

Replace

Code:
<xsl:copy-of select="current-group()"/>
with

Code:
<xsl:perform-sort select="current-group()">
  <xsl:sort select="local-name()"/>
  <xsl:sort select="."/>
</xsl:perform-sort>
__________________
Michael Kay
http://www.saxonica.com/
Author, XSLT 2.0 and XPath 2.0 Programmer\'s Reference
The Following User Says Thank You to mhkay For This Useful Post:
riteshdubey5 (July 2nd, 2012)
 
Old July 2nd, 2012, 05:13 AM
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Default

thanks, its working perfectly





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