Re: Accessing specific repetitive node

Ken,

Thanks for the response.  On this occasion, I was trying to access items 1,
3 & 5, but it could be any items and I do not know in advance, which is why
I need to select by position reference.

Any ideas why this might not be working.  It may help you to see the
response I just posted to David on this topic.

Thanks,

Mark

----- Original Message -----
From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>
To: "XSL-List" <XSL-List@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Friday, March 05, 2004 12:15 PM
Subject: Re:  Accessing specific repetitive node


> At 2004-03-05 11:47 +0000, Mark Williams wrote:
> >The XML looks like this
> >
> ><DATA>
> >     <Table ID="2">
> >         <Data>
> >             <Code>1</Code>
> >             <Code>2</Code>
> >             <Code>3</Code>
> >             <Code>4</Code>
> >             <Code>5</Code>
> >            </Data>
> >     </Table>
> ></DATA>
> >
> >I need to access the first 1, 3 and 5 items.
>
> Is it that you need to return all odd-numbered items?  Or do you need just
> the first, third and fifth items of all sets?
>
> >I have tried it in the
> >following ways:
> >
> >  <xsl:value-of select="DATA/Table[@ID='2']/Data/Code[1]" />  //this
works
> >okay
> >  <xsl:value-of select="DATA/Table[@ID='2']/Data/Code[position()=1]" />
> >//this also works okay
>
> Those two are equivalent: the first is the abbreviation of the second.
>
> <xsl:value-of> returns only a single item so you cannot use it to return
> multiple items.
>
> To access every odd item in the list you could use modulus arithmetic:
>
>    <xsl:for-each select="DATA/Table[@ID='2']/Data/Code[position() mod 2 =
1]">
>      <xsl:if test="position() > 1">, </xsl:if>
>      <xsl:value-of select="."/>
>    </xsl:for-each>
>
> I hope this helps.
>
> .............................. Ken
>
> --
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