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(_at_)CraneSoftwrights(_dot_)com>
To: "XSL-List" <XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Friday, March 05, 2004 12:15 PM
Subject: Re: [xsl] 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[(_at_)ID='2']/Data/Code[1]" />  //this

works

okay
 <xsl:value-of select="DATA/Table[(_at_)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[(_at_)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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