Re: xml naar xml transformeren

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#60

From: flor
Date: Fri, 29 Feb 2008 21:16

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flor wrote:
> 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?

opgelost, foutje gemaakt bij kopieren.
bedankt nogmaals!

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