CODEWITHSUNDEEP
xsltduplicates
user10821258
user10821258

Reputation: 13

Output only Duplicates in the final Text Output

I am trying to output only one line per unique value in my final text output after running an XML through an XSL stylesheet. In my research, I came upon the distinct-values function, but I'm unable to execute it the way that I want.

Here is my XML: 

<Library>
    <Book>
        <Code>1</Code>
        <Title>MANAGEMENT</Title>
    </Book>
    <Book>
        <Code>1</Code>
        <Title>MANAGEMENT</Title>
    </Book>
    <Book>
        <Code>1</Code>
        <Title>MANAGEMENT</Title>
    </Book>
    <Book>
        <Code>1</Code>
        <Title>MANAGEMENT</Title>
    </Book>
    <Book>
        <Code>1</Code>
        <Title>MANAGEMENT</Title>
    </Book>
    <Book>
        <Code>10</Code>
        <Title>MECHANICAL</Title>
    </Book>
    <Book>
        <Code>106</Code>
        <Title>TRANSPORTATION</Title>
    </Book>
</Library>

And here is my current XSL (incorrect):

<xsl:template match="Book">

    <xsl:value-of select="this:fixedOutput(Code)" />

    <xsl:value-of select="this:fixedOutput(Title)" />

    <xsl:value-of select="$linefeed" />

</xsl:template>

My output right now is:

1|MANAGEMENT| 1|MANAGEMENT| 1|MANAGEMENT| 1|MANAGEMENT| 1|MANAGEMENT| 10|MECHANICAL| 106|TRANSPORTATION|

But I want it to be this:

1|MANAGEMENT| 10|MECHANICAL| 106|TRANSPORTATION|

I'm not sure how to use the syntax of distinct values to get to where I need.

Upvotes: 0

Views: 37

Answers (1)

Vebbie
Vebbie

Reputation: 1695

An XSLT 1.0 solution that uses key and the generate-id() function (Muenchian grouping) to get distinct values :

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:key name="bookCode" match="/Library/Book/Code" use="." />

<xsl:template match="/">
    <xsl:for-each select="/Library/Book/Code[generate-id()
                                   = generate-id(key('bookCode',.)[1])]">

        <xsl:value-of select="this:fixedOutput(.)" />
        <xsl:value-of select="this:fixedOutput(../Title)" />
        <xsl:value-of select="$linefeed" />
    </xsl:for-each>
</xsl:template>

An XSLT 2.0 solution which uses xsl:for-each-group as @michael.hor257k said :

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="2.0" xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs">

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

<xsl:template match="Library">
    <xsl:for-each-group select="Book" group-by="concat(Code,Title)">
        <xsl:apply-templates select="." />
    </xsl:for-each-group>
</xsl:template>

<xsl:template match="Book">
    <xsl:value-of select="this:fixedOutput(Code)" />
    <xsl:value-of select="this:fixedOutput(Title)" />
    <xsl:value-of select="$linefeed" />
</xsl:template>

Note: As this:fixedOutput in your code doen't refer any namespace it has been used as it is.

Refer this: http://xsltransform.net/3MP2uBE

Upvotes: 1

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