Applying Muenchian grouping for a simple XML with XSLT

Question

How does Muenchian grouping work in details?

I have a simple XML document generated from a database:

<CLIENTS>
    <CLIENT>
       <NAME>John</NAME>
       <ACCOUNT_NUMBER>1424763562761</ACCOUNT_NUMBER>
       <LAST_USED>2012-10-03</LAST_USED>
       <AMOUNT>5000</AMOUNT>

    </CLIENT>
    <CLIENT>
       <NAME>John</NAME>
       <ACCOUNT_NUMBER>543667543732</ACCOUNT_NUMBER>
       <LAST_USED>2012-10-02</LAST_USED>
       <AMOUNT>10000</AMOUNT>
    </CLIENT>
       ...

I'd like to group by the name node. How can I the desired ouput to be the following?

<ClIENTS>
    <CLIENT>
        <NAME>John</NAME>
        <ACCOUNT>
           <ACCOUNT_NUMBER>1424763562761</ACCOUNT_NUMBER>
           <LAST_USED>2012-10-03</LAST_USED>
           <AMOUNT>5000</AMOUNT>
        </ACCOUNT>
        <ACCOUNT>
           <ACCOUNT_NUMBER>543667543732</ACCOUNT_NUMBER>
           <LAST_USED>2012-10-03</LAST_USED>
           <AMOUNT>10000</AMOUNT>
        </ACCOUNT>
       ....
</CLIENTS>

Answer 1

Read www.jenitennison.com/xslt/grouping/muenchian.xml, for a help with the code define a key

<xsl:key name="client-by-name" match="CLIENT" use="NAME"/>

then use templates as

<xsl:template match="@* | node()">
  <xsl:copy>
    <xsl:apply-templates select="@* | node()"/>
  </xsl:copy>
</xsl:template>


<xsl:template match="CLIENTS">
  <xsl:copy>
    <xsl:apply-templates select="CLIENT[generate-id() = generate-id(key('client-by-name', NAME)[1])]" mode="group"/>
  <xsl:copy>
</xsl:template>

<xsl:template match="CLIENT" mode="group">
  <xsl:copy>
    <xsl:copy-of select="NAME"/>
    <xsl:apply-templates select="key('client-by-name', NAME)"/>
  </xsl:copy>
</xsl:template>

<xsl:template match="CLIENT">
  <ACCOUNT>
    <xsl:apply-templates select="node()[not(self::NAME)]"/>
  </ACCOUNT>
</xsl:template>

[edit] If you want to use XSLT 2.0 then of course you don't need Muenchian grouping, instead you use

<xsl:template match="@* | node()">
  <xsl:copy>
    <xsl:apply-templates select="@* , node()"/>
  </xsl:copy>
</xsl:template>

<xsl:template match="CLIENTS">
  <xsl:copy>
    <xsl:for-each-group select="CLIENT" group-by="NAME">
      <CLIENT>
        <xsl:apply-templates select="NAME, current-group()"/>
      </CLIENT>
    </xsl:for-each-group>
  </xsl:copy>
</xsl:template>

<xsl:template match="CLIENT">
  <ACCOUNT>
    <xsl:apply-templates select="node() except NAME"/>
  </ACCOUNT>
</xsl:template>

Answer 2

Jeni Tennison breaks the steps required to perform Muenchian grouping here:

http://www.jenitennison.com/xslt/grouping/muenchian.html

Essentially, use XSLT to assign a key to the node, this key can be repeated if there's identical nodes in the document. The XSLT then goes through each key, and allows you to output the nodes with matching keys.

So, in Martin's answer, this line is creating a key for each CLIENT based on the content of the NAME node (remember if the NAME is the same for multiple clients, so will the key):

<xsl:key name="client-by-name" match="CLIENT" use="NAME"/>

You then want to go through all the keys and find the first instance of each (again using Martin's answer)

<xsl:apply-templates select="CLIENT[generate-id() = generate-id(key('client-by-name', NAME)[1])]" mode="group"/>

You then want to find all of the CLIENTS that match the key to be able to output their detail (again, Martins)

<xsl:apply-templates select="key('client-by-name', NAME)"/>

From here you'd need another template to output the CLIENT details