Re: xml naar xml transformeren

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#58

From: flor
Date: Fri, 29 Feb 2008 20:50

23 lines
2087 bytes

Suzan Foster wrote:
> Suzan Foster schreef:
> 
> Darn, ik blijf mezelf verbeteren..
> 
> ...
>  <xsl:template match="B[@N='BirthDate']">
>    <xsl:param name="date" select="substring-before(@V, ' ')"/>
>    <xsl:param name="year" select="substring-before($date, '-')"/>
>    <xsl:param name="month-and-day" select="substring-after($date, '-')"/>
>    <xsl:param name="month" select="substring-before($month-and-day, '-')"/>
>    <xsl:param name="day" select="substring-after($month-and-day, '-')"/>
>    <DateOfBirth>
>        <xsl:value-of select="concat($day, '/', $month, '/', $year)"/>
>    </DateOfBirth>
>  </xsl:template>
> ...
> 
> 
> Groeten, Su.

heel heel heel heel heel hartelijk bedankt!
werkt als een klok!

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