XSLT grouping / xsl:key problem

Hi,

I have a problem with grouping. My source XML has <record> elements that have a @name and a @group attribute. It looks something like this:

Expand|Select|Wrap|Line Numbers

 <root>

    <result>

        <record name="test1" group="group1"><value>1A</value></record>

        <record name="test2" group="group1"><value>1B</value></record>

        <record name="test3" group="group1"><value>1C</value></record>

        <record name="test1" group="group2"><value>2A</value></record>

        <record name="test3" group="group2"><value>2F</value></record>

        <record name="test4" group="group2"><value>2E</value></record>

        <record name="test2" group="group3"><value>3A</value></record>

        <record name="test3" group="group3"><value>3B</value></record>

        <record name="test5" group="group3"><value>3E</value></record>

    </result>

    <result>

        <record name="test1" group="group1"><value>1A</value></record>

        <record name="test2" group="group1"><value>1B</value></record>

        <record name="test3" group="group1"><value>1C</value></record>

        <record name="test8" group="group4"><value>4D</value></record>

        <record name="test9" group="group4"><value>4E</value></record>

    </result>

</root>

The goal is to have a html output, that shows the results as tables, one table per result which should look something like this:

---------------------------------------------
Result 1 | 1 | 2 | 3 |
---------------------------------------------
test1 | 1A | 2A | |
---------------------------------------------
test2 | 1B | | 3A |
---------------------------------------------
test3 | 1C | 2F | 3B |
---------------------------------------------
test4 | | 2E | |
---------------------------------------------
test5 | | | 3E |
---------------------------------------------

------------------------------------------------------
Result 2 | 1 | 2 | 3 | 4 |
------------------------------------------------------
test1 | 1A | | | |
------------------------------------------------------
test2 | 1B | | | |
------------------------------------------------------
test3 | 1C | | | |
------------------------------------------------------
test8 | | | | 4D |
------------------------------------------------------
test9 | | | | 4E |
------------------------------------------------------

For the numer of rows in the table I need to group the records by the @name attribute. I tried doing this with a <xsl:key>.

This is my XSLT stylesheet:

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 <xsl:key name="nameKey" match="record" use="@name"/>

<xsl:template match="/">

    <html>

       <head>

       </head>

       <body>

            <xsl:for-each select="root/result">

                <xsl:variable name="positionResult" select="position()"/>
 
                <xsl:element name="table">

                    <xsl:attribute name="border"><xsl:text>1</xsl:text></xsl:attribute>

                    <xsl:element name="thead">

                        <xsl:element name="tr">

                            <xsl:element name="td">

                                <xsl:text>Result </xsl:text><xsl:value-of select="$positionResult"/>

                            </xsl:element>

                            <xsl:element name="td"><xsl:text>1</xsl:text></xsl:element>

                            <xsl:element name="td"><xsl:text>2</xsl:text></xsl:element>

                            <xsl:element name="td"><xsl:text>3</xsl:text></xsl:element>

                            <xsl:element name="td"><xsl:text>4</xsl:text></xsl:element>

                            <xsl:element name="td"><xsl:text>5</xsl:text></xsl:element>

                        </xsl:element>

                    </xsl:element>

                    <xsl:element name="tbody">

                        <xsl:for-each select="record[count(. | key('nameKey', @name)[1]) = 1]">

                            <xsl:sort select="@name" order="ascending"/>

                            <xsl:variable name="names" select="@name"/>
 
                            <xsl:element name="tr">

                                <xsl:element name="td"><xsl:value-of select="$names"/></xsl:element>

                                <xsl:element name="td"><xsl:value-of select="//result[$positionResult]/record[@name = $names and @group = 'group1']/value"/><xsl:text>*</xsl:text></xsl:element>

                                <xsl:element name="td"><xsl:value-of select="//result[$positionResult]/record[@name = $names and @group = 'group2']/value"/><xsl:text>*</xsl:text></xsl:element>

                                <xsl:element name="td"><xsl:value-of select="//result[$positionResult]/record[@name = $names and @group = 'group3']/value"/><xsl:text>*</xsl:text></xsl:element>

                                <xsl:element name="td"><xsl:value-of select="//result[$positionResult]/record[@name = $names and @group = 'group4']/value"/><xsl:text>*</xsl:text></xsl:element>

                                <xsl:element name="td"><xsl:value-of select="//result[$positionResult]/record[@name = $names and @group = 'group5']/value"/><xsl:text>*</xsl:text></xsl:element>

                            </xsl:element>        
 
                        </xsl:for-each>

                    </xsl:element>

                </xsl:element>

            </xsl:for-each>

        </body>

    </html>

</xsl:template>

Somehow however the resulting HTML does not show me the rows for test1, test2 and test3 for the second result table. Instead it looks like this:

------------------------------------------------------
Result 2 | 1 | 2 | 3 | 4 |
------------------------------------------------------
test8 | | | | 4D |
------------------------------------------------------
test9 | | | | 4E |
------------------------------------------------------

I think this is because the <xsl:key> for these has already been used in the first iteration ??! There is no way to predict the number of result elements in the source XML so I can't use a different <xsl:key> for each iteration...

Is there some other way to group the elements ?

Thank you very much !!

Leira

Jun 25 '09 #1

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