XSLT Grouping and merging nodes - xslt

Question

Is it possible to group nodes and merge with a specified key? Some of my xml files have information about one person in two child nodes and I would like to find if nodes are about the same person and then marge it into one node
Input xml looks like this:
<db>
<next>
<name>John</name>
<surname>Smith</surname>
<number>304888</number>
<details>
<city>Westfield</city>
<zip-code>07090</zip-code>
<address>23 Victoria Street</address>
<phone>123456789</phone>
<fax/>
</details>
</next>
<next>
<name>John</name>
<surname>Smith</surname>
<number>304888</number>
<details>
<city>Westfield</city>
<zip-code>07090</zip-code>
<address>23 Victoria Street</address>
<phone>223344123</phone>
<fax>993456789</fax>
</details>
</next>
<next>
<name>John</name>
<surname>Smith</surname>
<number>113190</number>
<details>
<city>Richmond</city>
<zip-code>3121</zip-code>
<address>18 Seasame Street</address>
<phone>123456222</phone>
<fax/>
</details>
</next>
<next>
<name>John</name>
<surname>Smith</surname>
<number>113190</number>
<details>
<city>Richmond</city>
<zip-code>3133</zip-code>
<address>23 Baker Street</address>
<phone>113344123</phone>
<fax>133456789</fax>
</details>
</next>
</db>
'number' is an ID of the person and there can be more people with the same name and surename but with different 'number' value. Also, there can be the same person more than once (with the same value of 'name', 'surename' and 'number') but with different details. If the combination of 'city', 'zip-code' and 'address' are the same but 'phone' or/and 'fax' are differend then I would like to marge it.
And I want it to look like this:
<db>
<next>
<name>John</name>
<surname>Smith</surname>
<number>304888</number>
<details>
<city>Westfield</city>
<zip-code>07090</zip-code>
<address>23 Victoria Street</address>
<phone>123456789</phone>
<phone>223344123</phone>
<fax>993456789</fax>
</details>
</next>
<next>
<name>John</name>
<surname>Smith</surname>
<number>113190</number>
<details>
<city>Richmond</city>
<zip-code>3121</zip-code>
<address>18 Seasame Street</address>
<phone>123456222</phone>
<fax/>
</details>
</next>
<next>
<name>John</name>
<surname>Smith</surname>
<number>113190</number>
<details>
<city>Richmond</city>
<zip-code>3133</zip-code>
<address>23 Baker Street</address>
<phone>113344123</phone>
<fax>133456789</fax>
</details>
</next>
</db>
I have tried muenchian grouping but I don't know how to do it with phone and fax nodes.
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes" />
<xsl:key name="index" match="next" use="number" />
<xsl:key name="details-key" match="details" use="city,'_',zip-code,'_',address" />
<xsl:template match="#* | node()">
<xsl:copy>
<xsl:apply-templates select="#* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="db">
<xsl:copy>
<xsl:for-each select="//next[generate-id() = generate-id(key('index', number)[1])]">
<xsl:element name="name">
<xsl:value-of select="name"/>
</xsl:element>
<xsl:element name="surname">
<xsl:value-of select="surname"/>
</xsl:element>
<xsl:element name="number">
<xsl:value-of select="number"/>
</xsl:element>
<next>
<xsl:for-each select="//details[generate-id() = generate-id(key('details-key', concat(city,'_',zip-code,'_',address))[1])]">
<xsl:element name="city">
<xsl:value-of select="city"/>
</xsl:element>
<xsl:element name="zip-code">
<xsl:value-of select="zip-code"/>
</xsl:element>
<xsl:element name="address">
<xsl:value-of select="address"/>
</xsl:element>
</xsl:for-each>
</next>
</xsl:for-each>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>

Answer 1

First thing, your stylesheet will produce an error because the expression you used in:
<xsl:key name="details-key" match="details" use="city,'_',zip-code,'_',address" />
is not a valid expression in XPath 1.0.
If you want to group the details elements by the combined values of city, zip-code and address, you should define your key as:
<xsl:key name="details-key" match="details" use="concat(city, '_', zip-code, '_', address)" />
Now, if I go by your expected output instead of by your description, you actually want to create a group for each unique combination of number and address. So something along the lines of:
XSLT 1.0
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes" />
<xsl:key name="k1" match="next" use="concat(number, '_', details/city,'_', details/zip-code,'_', details/address)" />
<xsl:template match="db">
<xsl:copy>
<xsl:for-each select="next[generate-id() = generate-id(key('k1', concat(number, '_', details/city,'_', details/zip-code,'_', details/address))[1])]">
<next>
<xsl:copy-of select="name | surname | number"/>
<details>
<xsl:copy-of select="city | zip-code | address"/>
<xsl:variable name="group-details" select="key('k1', concat(number, '_', details/city,'_', details/zip-code,'_', details/address))/details" />
<xsl:copy-of select="$group-details/phone[text()] | $group-details/fax[text()]"/>
</details>
</next>
</xsl:for-each>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>