CODEWITHSUNDEEP
xsltxpathaxes
dirin
dirin

Reputation: 371

XPATH Axes - Issues

XML structure:

<units>
    <unit>
        <lesson>
            <name>Sample 1</name>
            <sections>
                <section type="IN">
                    <title> Sample Title 1 </title>
                </section>
            </sections>
        </lesson>
        <lesson>
            <name>Sample 2</name>
            <sections>
                <section type="OF">
                    <title> Sample Title 2 </title>
                </section>
            </sections>
        </lesson>
        <lesson>
            <name>Sample 3</name>
            <sections>
                <section type="IN">
                    <title> Sample Title 3</title>
                </section>
            </sections>
        </lesson>
        <lesson>
            <name>Sample 4</name>
            <sections>
                <section type="AS">
                    <title> Sample Title 4</title>
                </section>
            </sections>
        </lesson>
        <lesson>
            <name>Sample 5</name>
            <sections>
                <section type="IN">
                    <title> Sample Title 5</title>
                </section>
            </sections>
        </lesson>
    </unit>
</units>

My requirement is to get the values of title element and display as follows (Grouping similar data and display)

IN:
Sample Title 1
Sample Title 3
Sample Title 5
OF:
Sample Title 2
AS:
Sample Title 5

I have used following-sibling option to get the expected output. Since the XML structure is huge(I have pasted only the snippet), I cannot hard-code the path using ../../ and all in XSLT. Please help me in getting the expected output.

Upvotes: 2

Views: 320

Answers (2)

Michael Kay
Michael Kay

Reputation: 163575

Here is the solution in XSLT 2.0:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="text" />
    <xsl:template match="/">
        <xsl:for-each-group select="//section" group-by="@type">
           <xsl:value-of select="@type, ':&#xa;'" separator=""/>
           <xsl:value-of select="current-group()/title" separator="&#xA;" />
           <xsl:value-of select="'&#xa;'"/>
        </xsl:for-each-group>              
    </xsl:template>
</xsl:stylesheet>

Upvotes: 1

Wayne
Wayne

Reputation: 60424

It's better to solve this using grouping than either of the sibling axes:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="text" />
    <xsl:key name="bySectionType" match="section" use="@type" />
    <xsl:template match="/">
        <xsl:apply-templates select="units/unit/lesson/sections/section" />
    </xsl:template>
    <xsl:template match="section" />
    <xsl:template
        match="section[generate-id()=
                       generate-id(key('bySectionType', @type)[1])]">
        <xsl:value-of select="concat(@type, ':&#xA;')" />
        <xsl:apply-templates select="key('bySectionType', @type)" mode="out" />
    </xsl:template>
    <xsl:template match="section" mode="out">
        <xsl:value-of select="concat(normalize-space(title), '&#xA;')" />
    </xsl:template>
</xsl:stylesheet>

Output:

IN:
Sample Title 1
Sample Title 3
Sample Title 5
OF:
Sample Title 2
AS:
Sample Title 4

For completeness, the following stylesheet achieves the same result using the preceding and following axes:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="text" />
    <xsl:template match="/">
        <xsl:apply-templates select="units/unit/lesson/sections/section" />
    </xsl:template>
    <xsl:template match="section" />
    <xsl:template 
        match="section[not(preceding::section[@type=current()/@type])]">
        <xsl:value-of select="concat(@type, ':&#xA;')" />
        <xsl:apply-templates select=".|following::section[@type=current()/@type]"
            mode="out" />
    </xsl:template>
    <xsl:template match="section" mode="out">
        <xsl:value-of select="concat(normalize-space(title), '&#xA;')" />
    </xsl:template>
</xsl:stylesheet>

This is a far less efficient solution. The normal way to solve this is with the Muenchian Method for grouping, as shown above.

Upvotes: 3

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