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   xml schema error when deriving by extension

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  • To: xml dev <xml-dev@lists.xml.org>
  • Subject: xml schema error when deriving by extension
  • From: Anil Philip <goodnewsforyou@yahoo.com>
  • Date: Mon, 27 Feb 2006 14:55:10 -0800 (PST)
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  • In-reply-to: <43F357AC.1040507@allette.com.au>

Hello,
I am trying to derive from an existing type by
extension.
However I get the following error during validation in
XMLSpy:
"An 'all' model group is neither allowed in complex
type 'NodeType' nor in its extension
'RemoteNodeType'."

Where I have a tree of nodes consisting of

<xs:element name="AbcNode" type="NodeType"/>
<xs:complexType name="NodeType">
 <xs:all>
  <xs:element name="childNodes" type="ListOfNodes"/>
  <xs:element name="title" type="LocalSnippet"/>
  </xs:all>
 <xs:attribute name="id" type="xs:int"
use="required"/>
</xs:complexType>

I would like RemoteNodeType to extend NodeType and
contain two extra fields, url and id.

So I did:

<xs:element name="RemoteNode" type="RemoteNodeType"/>
  <xs:complexType name="RemoteNodeType">
    <xs:complexContent>
      <xs:extension base="NodeType">
         <xs:sequence>
 		<xs:element name="remoteNodeUrl"
type="RelativeURI"/>
		<xs:element name="remoteNodeId" type="xs:int"/>
       </xs:sequence>
      </xs:extension>
    </xs:complexContent>
  </xs:complexType>


  However I get the error as above. I would really
appreciate any help since I am stumped. I
  went through the W3C XML Schema Primer, but could
not fix it.
thanks,
Anil Philip




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