K
Kent B
Hello people,
Sorry if this is the wrong newsgroup. I had a look and couldn't find a
more appropriate one for XSD.
I have a complex type defined in its own schema as follows:
<xsd:complexType name="myType">
<xsd:complexContent>
<xsd:extension base="myBaseType">
<xsd:attribute name="parentId" type="xsd:string" use="optional"/>
</xsd:extension>
</xsd:complexContent>
</xsd:complexType>
I want to apply a keyRef so that the parentId attribute must refer to a
valid ID (defined elsewhere in another schema). However, as far as I
can
tell, XSD only allows you to specify a keyRef inside an element
declaration.
But I don't want to declare an element. Does anyone know how I can
declare a
keyRef without requiring the element?
Thanks,
Kent
Sorry if this is the wrong newsgroup. I had a look and couldn't find a
more appropriate one for XSD.
I have a complex type defined in its own schema as follows:
<xsd:complexType name="myType">
<xsd:complexContent>
<xsd:extension base="myBaseType">
<xsd:attribute name="parentId" type="xsd:string" use="optional"/>
</xsd:extension>
</xsd:complexContent>
</xsd:complexType>
I want to apply a keyRef so that the parentId attribute must refer to a
valid ID (defined elsewhere in another schema). However, as far as I
can
tell, XSD only allows you to specify a keyRef inside an element
declaration.
But I don't want to declare an element. Does anyone know how I can
declare a
keyRef without requiring the element?
Thanks,
Kent