how to get distinct values

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Palle Hansen 143 posts 396 karma points

Jun 13, 2016 @ 05:40

Templates, (partial) views and macros

Umbraco 6

Hi, I'm struggling with an xslt that is suppose to give me some distinct values that I'm gonna use as options in a select.

Here is what I've got:

<xsl:variable name="make" select="umbraco.library:Request('make')"/>        
<xsl:variable name="alldata" select="umbraco.library:GetXmlNodeById(1674)/Leasingcar"/>
<xsl:variable name="prods" select="msxml:node-set($alldata) [wwleasingform/MultiNodePicker/nodeId = $make]" />

<xsl:for-each select="msxml:node-set($prods) ">     
        <xsl:if test="generate-id() = generate-id(msxml:node-set($prods)[. = current()][1])">
            <option value="{wwbilmaerke}">
                <xsl:value-of select="umbraco.library:GetXmlNodeById(wwbilmaerke)/@nodeName"/>
            </option>
     </xsl:if> 
</xsl:for-each>

The for-each gives me : Mercedes, BMW, BMW, Audi and it should give me Mercedes, BMW, Audi

So I guess someting in the for-each isn't right.

Can someone see what I'm doing wrong?

Best regards Palle

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Jan Skovgaard 11280 posts 23678 karma points MVP 10x admin c-trib

Jun 13, 2016 @ 06:10

0

Hi Palle

I think you're missing a key in the above code - See the example by Tommy here http://blackpoint.dk/umbraco-workbench/xslt/grouping--distinct-values.aspx?p=3

Perhaps you can also find further inspiration in this thread https://our.umbraco.org/forum/developers/xslt/4482-xslt-distinct - Only be aware that since these examples where created the XML schema of Umbraco changed - So if you see "node/data[@alias ='something']" a lot, which was the old way to do the XSLT back then. Just mentioning so you don't get confused by it :)

Hope this helps!

/Jan

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Palle Hansen 143 posts 396 karma points

Jun 14, 2016 @ 14:03
1
Hi Jan.

Thanks, it' helped You were right. Got it to work with this:
```
<xsl:key name="lcars" match="leasingcar" use="."/>
```
Palle
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