xsd:enumeration inheritance fails

[aevans1108]

Greetings

I can't seem to inherit enumerated values from a globally defined type
in my XML schema. XmlSchema.Compile() doesn't like it.

Here's the schema.

<?xml version="1.0" encoding="UTF-8" ?>
<xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema"
elementFormDefault="qualified">
<xsd:simpleType name="MYTYPE">
<xsd:restriction base="xsd:string">
<xsd:enumeration value="Option1" />
<xsd:enumeration value="Option2" />
<xsd:enumeration value="Option3" />
</xsd:restriction>
</xsd:simpleType>
<xsd:element name="Element1" default="Option1">
<xsd:simpleType>
<xsd:restriction base="MYTYPE">
<xsd:enumeration value="Option4" />
<xsd:enumeration value="Option5" />
<xsd:enumeration value="Option6" />
</xsd:restriction>
</xsd:simpleType>
</xsd:element>
</xsd:schema>

Here's the C# source.

const String FILENAME = @"d:\work\schemaTest\schemaTest2.xsd";

static protected void ValidationCallbackOne(object sender,
ValidationEventArgs args)
{
Console.WriteLine( args.Message );
}

static void Main()
{
XmlSchema schema = XmlSchema.Read( new XmlTextReader( FILENAME ),
null );
schema.Compile(new ValidationEventHandler(ValidationCallbackOne));
}

Here's the output.

The Enumeration constraint failed. An error occurred at
file:///d:/work/schemaTest/schemaTest2.xsd, (13, 10).

Does anyone have any clues? As always, guesses are welcome.
Thanks!
Tony

 

[Stan Kitsis [MSFT]]

Hi Tony,

When you create new datatypes by restriction, you add constraints to the
possible values of the base type. Your base type "MYTYPE" only allows three
values - "Option1", "Option2", and "Option3". Values "Option4", "Option5",
and "Option6" are not allowed by this datatype. This is why your
restriction is invalid.

--
Stan Kitsis
Program Manager, XML Technologies
Microsoft Corporation

This posting is provided "AS IS" with no warranties, and confers no rights.

 

[aevans1108]

Thank you for your reply, Stan.

So what you're saying is that my second simpleType can further restrict
MYTYPE, but it cannot extend it using xsd:restrictiions, right?

T

 

[Stan Kitsis [MSFT]]

Right. If you want to extend it, you need to use xsd:extension

--
Stan Kitsis
Program Manager, XML Technologies
Microsoft Corporation

This posting is provided "AS IS" with no warranties, and confers no rights.

 

[aevans1108]

I've got this schema that a big company shipped with one of their
popular products and it is riddled with bugs like this. I should have
known that just because they're a big company doesn't mean that they're
infallible.

Okay, thanks for the info.

T

 

[aevans1108]

Would you demonstrate, Stan? I'm confused on the xsd:extension syntax.
I tried this:

<xsd:simpleType name="MYTYPE">
<xsd:restriction base="xsd:string">
<xsd:enumeration value="Option1" />
<xsd:enumeration value="Option2" />
<xsd:enumeration value="Option3" />
</xsd:restriction>
</xsd:simpleType>

<xsd:element name="Element1" default="Option1">
<xsd:simpleType>
<xsd:extension base="MYTYPE">
<xsd:enumeration value="Option4" />
<xsd:enumeration value="Option5" />
<xsd:enumeration value="Option6" />
</xsd:extension>
</xsd:simpleType>
</xsd:element>

.... but now the XmlSchema object won't even read it in. It tells me
that "extension is invalid in this context".

In case it's unclear, I want element1 to allow the values "Option1",
"Option2", "Option3", "Option4", "Option5", and "Option6".
Thanks
Tony

 

[aevans1108]

Hello again.

I changed the subject line to reflect this new, but related problem.

I can see that you can extend (ie: add values to) an xsd:enumeration
with an xsd:union, but XmlSchemaSimpleTypeUnion doesn't seem to expose
unioned types that are defined inline.

Sample schema:

<?xml version="1.0" encoding="UTF-8" ?>
<xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema"
elementFormDefault="qualified">

<xsd:simpleType name="MYTYPE">
<xsd:restriction base="xsd:string">
<xsd:enumeration value="Option1" />
</xsd:restriction>
</xsd:simpleType>

<xsd:element name="Element1" default="Option1">
<xsd:simpleType>
<xsd:union memberTypes="MYTYPE">
<xsd:simpleType>
<xsd:restriction base="xsd:string">
<xsd:enumeration value="Option2" />
</xsd:restriction>
</xsd:simpleType>
</xsd:union>
</xsd:simpleType>
</xsd:element>

</xsd:schema>

The following C# code only prints the name of the first member type in
the union (ie: the one defined globally). The second member type,
which is declared inline and nameless (the XmlSchema class requires it
to be nameless), doesn't seem to be accessible. Specifically, I want
to get at the facets of the inline type.

const String FILENAME = @"schemaTest3.xsd";

static protected void ValidationCallbackOne(object sender,
ValidationEventArgs args)
{
Console.WriteLine( args.Message.Substring(0,34) );
}

static void Main()
{
XmlSchema schema = XmlSchema.Read( new XmlTextReader( FILENAME ), null
);
schema.Compile(new ValidationEventHandler(ValidationCallbackOne));

XmlSchemaSimpleTypeUnion u =
(XmlSchemaSimpleTypeUnion)((XmlSchemaSimpleType)((XmlSchemaElement)schema.Items[1]).SchemaType).Content;

foreach( XmlQualifiedName entry in u.MemberTypes )
{
Console.WriteLine( entry.Name );
}

}

I tried giving the inline type a name, then using that name in the
memberTypes attribute of the xsd:union element, but the XmlSchema class
didn't like that. Yes, I know I could define the inline type globally,
but I don't have that much control over the schema.

Surely there must be a way to access this inline type. Can anyone help
me? As always, all suggestions are welcome.

Thanks
Tony

 

[aevans1108]

Argh! Figured it out.

Where MemberTypes contains the names of the types in the memberTypes
list, BaseTypes contains a list of the types defined inline.
<sigh>

Disregard.

Thanks
Tony

 

[Stan Kitsis [MSFT]]

Hi Tony,

Sorry for the late reply - I just got back from the holiday break. Glad you
figured it out.

--
Stan Kitsis
Program Manager, XML Technologies
Microsoft Corporation

This posting is provided "AS IS" with no warranties, and confers no rights.