Hi Raimund,
You could just replace the "l_name" by "concat(l_name, ' ', f_name)" within the
xsl:key, key() and xsl:sort.
BTW: Since the 2nd and 3rd for-each loops are identical you might consider
putting them into a separate template to avoid duplicated code.
Regards,
Patrik
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Von: Raimund Kammering raimund(_dot_)kammering(_at_)desy(_dot_)de
[mailto:xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com]
Gesendet: Dienstag, 7. März 2017 10:32
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Betreff: [xsl] sort on one key - duplicates
Hi Folks,
I have an XSL which pulls out all last names and creates a sorted list of
these. Works fine, but now I noticed that for duplicated last names only the
first person is created in the output! So I need to also take the first name
into account! But I’m lacking the idea to accomplish this! The XML, XSL looks
like this:
<list>
<subject name=“A">
<person>
<l_name>Doe</l_name>
<f_name>John</f_name>
<expert/>
<mail>john(_dot_)doe(_at_)somewhere(_dot_)com</mail>
</person>
<person>
<l_name>Doe</l_name>
<f_name>Johanna</f_name>
<expert/>
<mail>johanna(_dot_)doe(_at_)somewhere(_dot_)com</mail>
</person>
</subject>
<subject name=“B”>
<person>
<l_name>Mueller</l_name>
<f_name>Michael</f_name>
<expert/>
<mail>michael(_dot_)mueller(_at_)somewhere(_dot_)com</mail>
</person>
<person>
<l_name>Mueller</l_name>
<f_name>Joe</f_name>
<expert/>
<mail>joe(_dot_)mueller(_at_)somewhere(_dot_)com</mail>
</person>
</subject>
</list>
with the following XSL
<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; version="2.0">
<xsl:output indent="yes" method="xml"/>
<!-- Needed to avoid duplicates in "All" subject see: -->
<!-- http://www.jenitennison.com/xslt/grouping/muenchian.htm -->
<xsl:key name="l_names" match="subject/person" use="l_name"/>
<xsl:template match="/list">
<xsl:choose>
<xsl:when test="count(./*[@name='All']) = 0">
<xsl:element name="{name()}">
<xsl:element name="subject">
<xsl:attribute name="name">All</xsl:attribute>
<xsl:for-each select="subject/person[count(. | key('l_names',
l_name)[1]) = 1]">
<xsl:sort select="l_name"/>
<xsl:element name="person">
<xsl:copy-of select="l_name"/>
<xsl:copy-of select="f_name"/>
<xsl:element name="expert"/>
<xsl:copy-of select="mail"/>
</xsl:element>
</xsl:for-each>
</xsl:element>
<xsl:for-each select="subject">
<xsl:element name="{name()}">
<xsl:attribute name="name">
<xsl:value-of select="@*"/>
</xsl:attribute>
<xsl:for-each select="person">
<xsl:sort select="l_name"/>
<xsl:copy-of select="."/>
</xsl:for-each>
</xsl:element>
</xsl:for-each>
</xsl:element>
</xsl:when>
<xsl:otherwise>
<xsl:element name="{name()}">
<xsl:for-each select="subject">
<xsl:element name="{name()}">
<xsl:attribute name="name">
<xsl:value-of select="@*"/>
</xsl:attribute>
<xsl:for-each select="person">
<xsl:sort select="l_name"/>
<xsl:copy-of select="."/>
</xsl:for-each>
</xsl:element>
</xsl:for-each>
</xsl:element>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
<list>
<subject name=“All">
<person>
<l_name>Doe</l_name>
<f_name>John</f_name>
<expert/>
<mail>john(_dot_)doe(_at_)somewhere(_dot_)com</mail>
</person>
<person>
<l_name>Mueller</l_name>
<f_name>Michael</f_name>
<expert/>
<mail>michael(_dot_)mueller(_at_)somewhere(_dot_)com</mail>
</person>
</subject>
<subject name=“A">
<person>
<l_name>Doe</l_name>
<f_name>John</f_name>
<expert/>
<mail>john(_dot_)doe(_at_)somewhere(_dot_)com</mail>
</person>
<person>
<l_name>Doe</l_name>
<f_name>Johanna</f_name>
<expert/>
<mail>johanna(_dot_)doe(_at_)somewhere(_dot_)com</mail>
</person>
</subject>
<subject name=“B”>
<person>
<l_name>Mueller</l_name>
<f_name>Michael</f_name>
<expert/>
<mail>michael(_dot_)mueller(_at_)somewhere(_dot_)com</mail>
</person>
<person>
<l_name>Mueller</l_name>
<f_name>Joe</f_name>
<expert/>
<mail>joe(_dot_)mueller(_at_)somewhere(_dot_)com</mail>
</person>
</subject>
</list>
which produces the requested output except for that in the “All” list is only
filled the first one of a person which name is appearing multiple times!
So I need to in addition to the key on l_name need to also take the f_name into
account!
Raimund
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