Display unique values in xslt version 1

Question

Can someone help me plz to get a distinct values of attribute ID in node INDIC ?

Explication:

Im using xslt version 1.

This is how my xml looks like:

<Results><Result>
    <INDICS>
        <INDIC ID="I000207" LIB="A"/>
        <INDIC ID="I000208" LIB="B"/>
        <INDIC ID="I999999" LIB="C"/>
    </INDICS>
</Result>
<Result>
    <INDICS>
        <INDIC ID="I000207" LIB="A"/>
        <INDIC ID="I000208" LIB="B"/>
        <INDIC ID="I999999" LIB="C"/>
    </INDICS>
</Result>

this is how my template looks like, my template display all IDs

     <xsl:template match="/">
<xsl:variable name="INDICS" select="//INDIC"/>
     <PRODUITS>
     <INDICS>
     <xsl:for-each select="//Results/Result/INDICS/INDIC/@ID">
     <INDIC ID= "{.}" LIB="{$INDICS[@ID=current()]/@LIB}" />
     </xsl:for-each>
     </INDICS>
     </PRODUITS>
     </xsl:template>

the output that I get is:

<PRODUITS>
<INDICS>
<INDIC ID="I000207" LIB="A"/>
<INDIC ID="I000208" LIB="B"/>
<INDIC ID="I999999" LIB="C"/>
<INDIC ID="I000207" LIB="A"/>
<INDIC ID="I000208" LIB="B"/>
<INDIC ID="I999999" LIB="C"/>
 <INDICS/>
</PRODUITS>

the output that Im trying to get is:

<PRODUITS>
<INDICS>
<INDIC ID="I000207" LIB="A"/>
<INDIC ID="I000208" LIB="B"/>
<INDIC ID="I999999" LIB="C"/>
 <INDICS/>
</PRODUITS>

thanks alot.

There are already many answer like this. Next time please use a [search like this](https://stackoverflow.com/search?q=%5Bxslt%5D+unique) before asking. — zx485, Oct 26 '18 at 14:35
It's from the usage guidelines for this site. See the second paragraph of [How do I ask a good question?](https://stackoverflow.com/help/how-to-ask). — zx485, Oct 26 '18 at 14:51

score 2 · Answer 1 · edited Oct 26 '18 at 15:26

2

Another possibility used in the linked questions is using a xsl:key and the generate-id() function. It is a variation of the so called Muenchian Grouping.

<xsl:key name="values" match="INDIC" use="@ID"/>

<xsl:template match="/">
    <PRODUITS>
        <INDICS>
            <xsl:for-each select="//Results/Result/INDICS/INDIC[generate-id() = generate-id(key('values',@ID)[1])]">
                <INDIC ID= "{@ID}" LIB="{@LIB}" />
            </xsl:for-each>
        </INDICS>
    </PRODUITS>
</xsl:template>

edited Oct 26 '18 at 15:26

Tim C

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answered Oct 26 '18 at 15:23

zx485

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Sacamoto - It is well worth learning about this techique, as is it is much more efficient with larger data sources. See http://www.jenitennison.com/xslt/grouping/muenchian.html – Tim C Oct 26 '18 at 15:29

score 0 · Accepted Answer · answered Oct 26 '18 at 15:15

You can do like this, hope this code help you to solve your problem:

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

    <xsl:output indent="yes"/>

    <xsl:template match="/">
        <Results>
            <Result>
                <INDICS>
                    <xsl:for-each select="//INDIC">
                        <xsl:variable name="c-id" select="@ID"/>
                        <xsl:choose>
                            <xsl:when test="$c-id = ancestor::Result/following-sibling::Result/INDICS/INDIC/@ID"/>
                            <xsl:otherwise>
                                <INDIC ID= "{@ID}" LIB="{@LIB}" />
                            </xsl:otherwise>
                        </xsl:choose>
                    </xsl:for-each>
                </INDICS>
            </Result>
        </Results>
    </xsl:template>

</xsl:stylesheet>

that's exaclty what I was looking for ! thanks Alot ! I really appreciate it :) — Sacamoto, Oct 26 '18 at 15:20

Display unique values in xslt version 1

2 Answers2