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XSD keyRef

Hello,

I have a complex type defined in its own schema as follows:

<xsd:complexTyp e name="myType">
<xsd:complexCon tent>
<xsd:extensio n base="myBaseTyp e">
<xsd:attribut e name="parentId" type="xsd:strin g" use="optional"/>
</xsd:extension>
</xsd:complexCont ent>
</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
Nov 15 '05 #1
2 1815
You can't. key/kerefs must be declared on an element. It could be either a
global element or a local element.

--
Stan Kitsis
Program Manager, XML Technologies
Microsoft Corporation

This posting is provided "AS IS" with no warranties, and confers no rights.
Use of included script samples are subject to the terms specified at
http://www.microsoft.com/info/cpyright.htm

"Kent Boogaart" <ke****@interno de.on.net> wrote in message
news:uH******** ******@TK2MSFTN GP12.phx.gbl...
Hello,

I have a complex type defined in its own schema as follows:

<xsd:complexTyp e name="myType">
<xsd:complexCon tent>
<xsd:extensio n base="myBaseTyp e">
<xsd:attribut e name="parentId" type="xsd:strin g" use="optional"/>
</xsd:extension>
</xsd:complexCont ent>
</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

Nov 15 '05 #2
Thanks for the reply. Is this a known flaw in XSD? Will something be done
about it? It seems a bit limiting to me. I have a similar problem trying to
declare uniqueness on a complex type's outer element's children.

Thanks again,
Kent

"Stan Kitsis [MSFT]" <sk***@microsof t.com> wrote in message
news:43******** @news.microsoft .com...
You can't. key/kerefs must be declared on an element. It could be either
a global element or a local element.

--
Stan Kitsis
Program Manager, XML Technologies
Microsoft Corporation

This posting is provided "AS IS" with no warranties, and confers no
rights. Use of included script samples are subject to the terms specified
at http://www.microsoft.com/info/cpyright.htm

"Kent Boogaart" <ke****@interno de.on.net> wrote in message
news:uH******** ******@TK2MSFTN GP12.phx.gbl...
Hello,

I have a complex type defined in its own schema as follows:

<xsd:complexTyp e name="myType">
<xsd:complexCon tent>
<xsd:extensio n base="myBaseTyp e">
<xsd:attribut e name="parentId" type="xsd:strin g" use="optional"/>
</xsd:extension>
</xsd:complexCont ent>
</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


Nov 15 '05 #3

This thread has been closed and replies have been disabled. Please start a new discussion.

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