Re: xml naar xml transformeren

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#59

From: flor
Date: Fri, 29 Feb 2008 21:03

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flor wrote:
> 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!

oei! te vroeg victorie gekraaid...
:(
blijkbaar neemt hij de tags niet allemaal over
het resultaat bevat niet de tags <Subject> en <Name>
hoe kan dat?

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