Re: [xsl] adding unique values to a drop-down list

["Thomas B. Passin" <tpassin@xxxxxxxxxxxx>]

[Katie McNally]

> I have not been able to correctly fill the drop-down with brokers.  The
> drop-down must contain the broker name/id combinations listed
alphabetically
> with no duplicates.
>
>

The suggestion you were trying out contained

<xsl:for-each select="Loan[not(.=preceding-sibling::*)

This would prevent unique entries if the items were sorted, but I'm not
clear they are.  There is at least one standard way to get a unique list,
using keys - it's in the various FAQs.  You have create a key that returns
the items.  Then you  set a variable equal to what the key returns,,
provided it is the first if there are any duplicates,  and then you have a
unique list.

This approach does not depend on any assumptions about the original ordering
of the items.

In this case you want to get a unique list of brokers based on their Id
values.  Here is a stylesheet that picks out your unique brokers - you can
modify it to produce the html that you want (I wrapped your example XML in a
"data" element"):

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>

<xsl:key name='brokers' match='Broker'  use='Id'/>

<xsl:variable name='items'
 select='/data/Loan/BrokerSet/Broker[
    generate-id(Id)=generate-id(key("brokers",Id)[1]/Id)]'/>

<xsl:template match="/">
<results>
<xsl:for-each select='$items'><!--This is a unique list -->
 <item><xsl:value-of select='Id'/></item>
</xsl:for-each>
</results>
</xsl:template>
</xsl:stylesheet>

Cheers,

Tom P


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