Question

How do I group repeating compositor content with xml stylesheet?

Asked by: Henrik_Nyholm

Dear Experts!

I have an XML file with content which might be described as an repeating sequence with elements all optional.

I want to transform this xml into a nice grouped xml structure.

How do I do?

Regards,

Input XML Schema:
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" elementFormDefault="qualified" attributeFormDefault="unqualified">
	<xs:element name="rootcanal">
		<xs:complexType>
			<xs:sequence maxOccurs="unbounded">
				<xs:element name="one" minOccurs="0"/>
				<xs:element name="two" minOccurs="0"/>
				<xs:element name="three" minOccurs="0"/>
			</xs:sequence>
		</xs:complexType>
	</xs:element>
</xs:schema>
 
Input XML:
<rootcanal xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="repeatComp.xsd">
	<one>1</one>
	<two>1</two>
	<three>1</three>
	<one>2</one>
	<three>2</three>
	<one>3</one>
	<two>3</two>
	<two>4</two>
	<three>4</three>
</rootcanal>
 
Output schema:
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" elementFormDefault="qualified" attributeFormDefault="unqualified">
	<xs:element name="rootcanal">
		<xs:complexType>
			<xs:sequence>
				<xs:element name="group" maxOccurs="unbounded">
					<xs:complexType>
						<xs:sequence>
							<xs:element name="one" minOccurs="0"/>
							<xs:element name="two" minOccurs="0"/>
							<xs:element name="three" minOccurs="0"/>
						</xs:sequence>
					</xs:complexType>
				</xs:element>
			</xs:sequence>
		</xs:complexType>
	</xs:element>
</xs:schema>
 
Output XML:
<rootcanal xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="repeatGroup.xsd">
	<group>
		<one>1</one>
		<two>1</two>
		<three>1</three>
	</group>
	<group>
		<one>2</one>
		<three>2</three>
	</group>
	<group>
		<one>3</one>
		<two>3</two>
	</group>
	<group>
		<two>4</two>
		<three>4</three>
	</group>
</rootcanal>
                                  
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Asked On
2009-10-31 at 14:00:06ID24861131
Tags

XML

,

XSLT

Topics

Extensible Markup Language (XML)

,

Extensible Stylesheet Language Transformation (XSLT)

Participating Experts
1
Points
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Comments
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    Answers

     

    by: GertonePosted on 2009-11-01 at 02:20:28ID: 25713192

    Here is all the XSLT you need

    <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:key name="rt" match="rootcanal/*" use="text()"/>
        <xsl:strip-space elements="*"/>
        <xsl:output indent="yes"/>
        <xsl:template match="rootcanal">
            <xsl:copy>
                <xsl:for-each select="*[generate-id() = generate-id(key('rt', text())[1])]">
                    <xsl:element name="group">
                        <xsl:copy-of select="key('rt', text())"/>
                    </xsl:element>
                </xsl:for-each>
            </xsl:copy>
        </xsl:template>
    </xsl:stylesheet>
                                                  
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    by: GertonePosted on 2009-11-01 at 02:21:15ID: 25713195

    I used muenchian grouping,
    which is very well explained here
    http://www.jenitennison.com/xslt/grouping/muenchian.xml

     

    by: Henrik_NyholmPosted on 2009-11-01 at 02:47:39ID: 25713250

    Dear Gertone!

    Thanks for reply.

    Anyway, your solution does work on the sample data I gave, but it does not work in general if you change the values of the elements.

    Your solution focus on grouping the values rather than taking the structure into account.

    What I am looking for is a solution that will work with the XML Schema structure despite of the actual data content.

    I hope you can come up with the brilliant solution on this one too...?

    Thanks in advance,
    in anticipation,
    best regards

     

    by: GertonePosted on 2009-11-01 at 05:30:32ID: 25713641

    Well, I didn't look at the schemata actually,
    just at your input data,
    yeah, well that doesn't work actually.
    I think I now understand what you need,
    you could have been a bit more explicit about that,

    anyway, I have to think about this, not so straightforward I am afraid
    CAn you use XSLT2?

     

    by: GertonePosted on 2009-11-01 at 06:11:08ID: 25713724

    Now that I understand what you want,
    I would solve this
    - in XSLT2
    - or in a two pass XSLT1

    Since it is only three elements that need grouping, you could test all conditions,
    so an approach could be to iterate over all the elements that can start a new group
    That is: one, two that doesn't have a direct preceding sibling one and three that doesn't have a direct preceding sibling one or two
    If you need an implementation in XSLT1 having more grouping elements, eg. <four> and <five> I recommend you two make this a two step... I can imagine some more elegant approaches

    <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
        <xsl:key name="rt" match="rootcanal/*" use="text()"/>
        <xsl:strip-space elements="*"/>
        <xsl:output indent="yes"/>
        <xsl:template match="rootcanal">
            <xsl:copy>
                <xsl:for-each select="one | two[not(preceding-sibling::*[1]/self::one)] | three[not(preceding-sibling::*[1]/self::one)][not(preceding-sibling::*[1]/self::two)]">
                    <xsl:element name="group">
                        <xsl:copy-of select="."/>
                        <xsl:if test="self::one">
                            <xsl:if test="following-sibling::*[1]/self::two">
                                <xsl:copy-of select="following-sibling::*[1]"/>
                                <xsl:if test="following-sibling::*[2]/self::three">
                                    <xsl:copy-of select="following-sibling::*[2]"/>
                                </xsl:if>
                            </xsl:if>
                            <xsl:if test="following-sibling::*[1]/self::three">
                                <xsl:copy-of select="following-sibling::*[1]"/>
                            </xsl:if>
                        </xsl:if>
                        <xsl:if test="self::two">
                            <xsl:if test="following-sibling::*[1]/self::three">
                                <xsl:copy-of select="following-sibling::*[1]"/>
                            </xsl:if>
                        </xsl:if>
                    </xsl:element>
                </xsl:for-each>
            </xsl:copy>
        </xsl:template>
    </xsl:stylesheet>
    
                                                  
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    by: Henrik_NyholmPosted on 2009-11-01 at 07:39:55ID: 25714024

    Dear Gertone!

    Thanks for reply.

    Yes, I should have made it more clear, that my focus was on the Schema construct rather than the data...

    Yes, I need a solution for XSLT 1

    And yes, I am looking for a generic solution.

    How would you do this in two step in XSLT 1?

    Actually constructs like this is quite common in XHTML, but a mess when you want to (re)structure it for database tables or the like...

    Thanks a lot.

    Regards,

     

    by: GertonePosted on 2009-11-02 at 00:46:30ID: 25717846

    In a more general way
    I would walk through the child elements recursively,
    passing a parameter containing the "history" and a parameter for current group
    based on the current element name and the history you could either keep the group or make a new group
    and you copy the element but add an attribute with the group
    This way you have selected for each node in which group it belongs

    The second pass would be a simple muenchian based on the group attribute

    You have now seperated the actual grouping (which changes the structure of the document and could give you some issues with nesting)
    from the indicator step that decides which group a node belongs too

    In XSLT2, you could easily do that in one step because you can list the new-group-triggers in a for-each-group statement

    If you are dealing with nesting XHTML, you need to take into account that a h2 nests deeper than a h1... which is yet another issue.
    Doing that in XSLT1 would give you a pipeline of processes to be comfortable. It is a piece of cake in XSLT2
    If you are realy nesting HTML, I recommend that you evaluate using XSLT2

     

    by: GertonePosted on 2009-11-02 at 02:04:26ID: 25718179

    Well, actually I am happy with the elegance and the robustness of the approach,
    I hope you are too
    It is not too obvious a task though

    Note that grouping is a very tricky business in XSLT1
    I have done my bit of XHTML processing and Word-XML processing
    and I must say that that is even tricky in XSLT2 :-(

    <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
        <xsl:strip-space elements="*"/>
        <xsl:output indent="yes"/>
        <xsl:template match="rootcanal">
            <xsl:copy>
                <xsl:if test="*">
                    <xsl:call-template name="process-node">
                        <xsl:with-param name="node" select="*[1]"/>
                        <xsl:with-param name="history" select="''"/>
                        <xsl:with-param name="running-group" select="1"/>
                    </xsl:call-template>
                </xsl:if>
            </xsl:copy>
        </xsl:template>
        
        <xsl:template name="process-node">
            <xsl:param name="node"/>
            <xsl:param name="history"/>
            <xsl:param name="running-group"/>
            <!-- if the name of this node is already in the history, start an new group, groups are numbered sequentially, first group is 1 -->
             <xsl:variable name="this-group">
                <xsl:choose>
                    <xsl:when test="contains($history, name($node))">
                        <xsl:value-of select="$running-group + 1"/>
                    </xsl:when>
                    <xsl:otherwise>
                        <xsl:value-of select="$running-group"/>
                    </xsl:otherwise>
                </xsl:choose>
             </xsl:variable>
            <!-- the history is virtual in that it is based on the current node name only, builds up the sequence from the schema -->
            <xsl:variable name="next-history">
                <xsl:choose>
                    <xsl:when test="$node/self::one"><xsl:text>-one-</xsl:text></xsl:when>
                    <xsl:when test="$node/self::two"><xsl:text>-one-two-</xsl:text></xsl:when>
                    <xsl:when test="$node/self::three"><xsl:text>-one-two-three-</xsl:text></xsl:when>
                </xsl:choose>
            </xsl:variable>
            <!-- the element is created again, but having a current-group attribute -->
            <xsl:element name="{name($node)}">
                <xsl:copy-of select="$node/@*"/>
                <xsl:attribute name="group"><xsl:value-of select="$this-group"/></xsl:attribute>
                <xsl:copy-of select="$node/node()"/>
            </xsl:element>
            <!-- and now we visit the next node -->
            <xsl:if test="$node/following-sibling::*">
                <xsl:call-template name="process-node">
                    <xsl:with-param name="node" select="$node/following-sibling::*[1]"/>
                    <xsl:with-param name="history" select="$next-history"/>
                    <xsl:with-param name="running-group" select="$this-group"/>
                </xsl:call-template>
            </xsl:if>
       </xsl:template>
            
     
    </xsl:stylesheet>
                                                  
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    by: GertonePosted on 2009-11-02 at 02:06:09ID: 25718183

    I forgot to add one line in my last comment...

    so it is not shamefull to make it multipass. I am even convinced that multi-pass is often a best practice when doing complex grouping

     

    by: Henrik_NyholmPosted on 2009-11-02 at 03:28:19ID: 31648444

    Brilliant

    20120131-EE-VQP-002

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