What is wrong with extending an XML Schema type using xs:all?

<?xml version="1.0" encoding="utf-8"?>
<xs:schema xmlns="http://tempuri.org/ServiceDescription.xsd" xmlns:mstns="http://tempuri.org/ServiceDescription.xsd" xmlns:xs="http://www.w3.org/2001/XMLSchema" targetNamespace="http://tempuri.org/ServiceDescription.xsd" elementFormDefault="qualified" id="ServiceDescription">
    <xs:element name="Template">
        <xs:complexType>
            <xs:complexContent>                                     
                <xs:extension base="ServiceType">
                    <xs:all>
                        <xs:element name="TemplateCode" type="xs:string"/>
                    </xs:all>
                </xs:extension>
            </xs:complexContent>
        </x开发者_如何学JAVAs:complexType>
    </xs:element>
    <xs:complexType name="ServiceType">
        <xs:all>
            <xs:element name="ServiceCode" type="xs:string"/>
        </xs:all>
    </xs:complexType>
</xs:schema>

When I try to save this in XMLSpy it tells me

An 'all' model group is neither allowed in complex type definition 'mstns:ServiceType' nor in its extension '{anonymous}'.

Clicking Details gives a link to a paragraph in XML Schema specification which I do not understand.

Added: Ah, yes, forgot to mention - the line of error is this one:

<xs:element name="TemplateCode" type="xs:string"/>

The problem is you can't have all if you're extending another type. As far as XML knows the parent type might have a sequence model and since XML forbids putting an all group inside of a sequence group (since that would destroy the sequence group's ordering) then XML also forbids putting an all group in an extension of a complex type. You could use sequence instead of all for both though and you'd be fine.