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RE: adding unique values to a drop-down list
- From: "Katie McNally" <kmcnally9 at hotmail dot com>
- To: xsl-list at lists dot mulberrytech dot com
- Date: Tue, 04 Dec 2001 14:50:35 -0600
- Subject: RE: [xsl] adding unique values to a drop-down list
- Reply-to: xsl-list at lists dot mulberrytech dot com
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 first time I followed your suggestion but the drop down did not contain
any brokers. My for-each statement:
<xsl:for-each select="Loan[not(.=preceding-sibling::*) and
string(BrokerSet/Broker/Name) and string(BrokerSet/Broker/Id)]">
<option value="{BrokerSet/Broker/Id}"><xsl:value-of
select="BrokerSet/Broker/Name"/> - <xsl:value-of
select="BrokerSet/Broker/Id"/></option>
</xsl:for-each>
I modified the for-each. The drop-down contained brokers, but they were not
listed alphabetically, and duplicates were present. My for-each:
<xsl:for-each select="Loan/BrokerSet/Broker[not(.=preceding-sibling::*) and
string(Name) and string(Id)]">
<option value="{Id}"><xsl:value-of select="Name"/> - <xsl:value-of
select="Id"/></option>
</xsl:for-each>
I added sort statements. The drop-down contained brokers, listed
alphabetically, but duplicates were present. My for-each:
<xsl:for-each select="Loan/BrokerSet/Broker[not(.=preceding-sibling::*) and
string(Name) and string(Id)]">
<xsl:sort select="Name"/>
<xsl:sort select="Id"/>
<option value="{Id}"><xsl:value-of select="Name"/> - <xsl:value-of
select="Id"/></option>
</xsl:for-each>
I cannot figure out how NOT to display duplicates. Any suggestions?
>From: "kfricovsky" <kfricovsky@fusebox.com>
>Reply-To: xsl-list@lists.mulberrytech.com
>To: <xsl-list@lists.mulberrytech.com>
>Subject: RE: [xsl] adding unique values to a drop-down list
>Date: Tue, 4 Dec 2001 12:18:19 -0500
>
>This will work for you. I used <funds> as my root element.
>
><xsl:stylesheet version="1.0"
>xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
> <xsl:output method="html"/>
> <xsl:template match="fund">
> <form action="">
> <select name="FOO">
> <xsl:for-each select="Loan[not(.=preceding-sibling::*)
>and string(BrokerSet/Broker/Name) and string(BrokerSet/Broker/Name)]">
> <option
>value="{BrokerSet/Broker/Id}"><xsl:value-of
>select="BrokerSet/Broker/Name" /> - <xsl:value-of
>select="BrokerSet/Broker/Id" /></option>
> </xsl:for-each>
> </select>
> </form>
> </xsl:template>
></xsl:stylesheet>
>
>-Kevin Fricovsky
>
>
>-----Original Message-----
>From: owner-xsl-list@lists.mulberrytech.com
>[mailto:owner-xsl-list@lists.mulberrytech.com] On Behalf Of Katie
>McNally
>Sent: Tuesday, December 04, 2001 11:45 AM
>To: xsl-list@lists.mulberrytech.com
>Subject: [xsl] adding unique values to a drop-down list
>
>
>I have xml that contains loans and their data (which includes broker
>name
>and id).
>
>I need to add a drop-down of brokers to my page. Each selection in the
>list
>box needs to display "Broker Name - Broker Id" and the value of each
>selection must be set to the Broker Id. These selections must be listed
>
>alphabetically, with only unique values displayed (no duplicates).
>
>For example:
>
>Example XML:
>
><Loan>
> <BrokerSet>
> <Broker>
> <Name>ABC Broker</Name>
> <Id>123456</Id>
> </Broker>
> </BrokerSet>
></Loan>
><Loan>
> <BrokerSet>
> <Broker>
> <Name>LMN Broker</Name>
> <Id>345678</Id>
> </Broker>
> </BrokerSet>
></Loan>
><Loan>
> <BrokerSet>
> <Broker>
> <Name/>
> <Id/>
> </Broker>
> </BrokerSet>
></Loan>
><Loan>
> <BrokerSet>
> <Broker>
> <Name>LMN Broker</Name>
> <Id>345678</Id>
> </Broker>
> </BrokerSet>
></Loan>
><Loan>
> <BrokerSet>
> <Broker>
> <Name>ABC Broker</Name>
> <Id>999999</Id>
> </Broker>
> </BrokerSet>
></Loan>
>
>For the xml listed above, the list box should contain the following
>values: All Brokers ABC Broker - 123456 ABC Broker - 999999 LMN Broker -
>345678
>
>How do I fill the list box with only unique selections which consist of
>broker name/broker id combinations?
>How do I handle loans that do not have broker name or id data?
>
>Thanks,
>Katie
>
>
>
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