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Re: Quick question regarding XPath


thunderbird xpath
Murali Mani wrote:

> is it possible to say in XPath
> 
> a//@b
> 
> is the above equivalent to:
> 
> (a/@b | a//*/@b)
> 

Yes, it is.  The step "//" means "starting from where you are, take 
descendents at any nesting level".  By default it starts at the root of 
the document.  As  an example, with this xml document -

<a b='1'>
   <a b='1.1'>
   	<a b='1.1.1'/>
   	<a b='1.1.2'/>
   </a>
</a>

this template gets all the "b" attribute values -

xsl:template match="/">
<results>
    <xsl:apply-templates select='a//@b'/>
</results>
</xsl:template>

Cheers,

Tom P

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