Re: position() returns doubled result - what do i miss

On Mon, 31 Mar 2003, jian wrote:

> i have :
>
> <?xml version="1.0" encoding="UTF-8" ?>
> <root>
>       <item>item1</item>
>       <item>item2</item>
>       <item>item3</item>
>       <item>item4</item>
>       <item>item5</item>
>       <item>item6</item>
>       <item>item7</item>
>       <item>item8</item>
>       <item>item9</item>
> </root>
>
>
>
> to which i apply:
>
> <?xml version="1.0" encoding="UTF-8" ?>
> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
>
> <xsl:output method="html" version="4.01" encoding="UTF-8"/>
>
>  <xsl:template match="/">

// here is your error, should be

<xsl:apply-templates select="root/item"/>

>    <xsl:apply-templates match="root/item"/>
>  </xsl:template>
>
>  <xsl:template match="item">

// replaced this with:

<xsl:template match="//item">


>    <xsl:value-of select="position()"/>
>    <xsl:value-of select="."/>
>  </xsl:template>
>
> </xsl:stylesheet>
>
>

XSLT:
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">

<xsl:output method="html" version="4.01" encoding="UTF-8"/>

 <xsl:template match="/">
   <xsl:apply-templates select="root/item"/>
 </xsl:template>

 <xsl:template match="//item">
        <p>
        <xsl:value-of select="position()"/>
        <xsl:value-of select="."/></p>
 </xsl:template>

</xsl:stylesheet>


<p>1item1</p>
<p>2item2</p>
<p>3item3</p>
<p>4item4</p>
<p>5item5</p>
<p>6item6</p>
<p>7item7</p>
<p>8item8</p>
<p>9item9</p>


Cheers,
Jarkko


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Jarkko Moilanen          "Erehtyminen on inhimillista,
Researcher                mutta todella suuret mokat
jm60697@xxxxxx            vaativat tietokoneen käyttöä."
www.uta.fi/~jm60697
GSM: +358 50 3766 927
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