[XML-DEV Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message]

Re: Quick question regarding XPath

• To: xml-dev@l...
• Subject: Re: Quick question regarding XPath
• From: "Thomas B. Passin" <tpassin@c...>
• Date: Wed, 25 Feb 2004 09:36:44 -0500
• In-reply-to: <Pine.GSO.4.58.0402250549490.17412@p...>
• References: <Pine.GSO.4.58.0402250549490.17412@p...>
• User-agent: Mozilla Thunderbird 0.5 (Windows/20040207)

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

• Follow-Ups:
  • Re: Quick question regarding XPath
    • From: "Thomas B. Passin" <tpassin@c...>

• References:
  • Quick question regarding XPath
    • From: Murali Mani <mani@C...>

• Prev by Date: Re: Quick question regarding XPath
• Next by Date: Re: Quick question regarding XPath
• Previous by thread: RE: Quick question regarding XPath
• Next by thread: Re: Quick question regarding XPath
• Index(es):
  • Date
  • Thread