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Re: How to copy only the really existing <choice> bra

Subject: Re: How to copy only the really existing <choice> branch/element? "value-of-if-exist"
From: David Carlisle <davidc@xxxxxxxxx>
Date: Wed, 7 Oct 2009 13:24:41 +0100
Re:  How to copy only the really existing <choice>  bra
>   If I simply write:
> 
> <xsl:value-of select=".../aaa"/>
> <xsl:value-of select=".../bbb"/>
> <xsl:value-of select=".../ccc"/>

> then ALL three elements were copied.
As Martin just said, xsl:value-of never copies elements.

I suspect you dod not simply write that but instead wrote

If I simply write:

<aaa>
<xsl:value-of select=".../aaa"/>
</aaa>
<bbb>
<xsl:value-of select=".../bbb"/>
</bbb>
<ccc>
<xsl:value-of select=".../ccc"/>
</ccc>

in which case the elements are not made by the value-of, they are made
by the literal result elements.

don't do that, just do

<xsl:copy-of select=".../top/*"/>

which will copy the child element of top which according to your schema
fragment must be aaa, bbb or ccc.

If the content of top had been more complicated but you just wanted
aaa,bbb or ccc then you could do



<xsl:copy-of select=".../top/(aaa|bbb|ccc)"/>


or (if you are still using xslt 1)


<xsl:copy-of select=".../top/*[self::aa or self::bbb or self::ccc]"/>


David

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