Using XSDs with includes

Richard Barraclough picture Richard Barraclough · Apr 12, 2012 · Viewed 9.9k times · Source

Here is an XSD:

<?xml version="1.0"?>
<xsd:schema 
elementFormDefault='unqualified' 
attributeFormDefault='unqualified' 
xmlns:xsd='http://www.w3.org/2001/XMLSchema' 
>

  <xsd:simpleType name='TheSimpleType'>
    <xsd:restriction base='xsd:string' />
  </xsd:simpleType>
</xsd:schema>

Here is a second XSD that includes the one above:

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

  <xsd:include schemaLocation='Include.xsd' />

  <xsd:element name = "TheElement" >
  <xsd:complexType>
  <xsd:attribute name="Code" type="TheSimpleType" use="required"/>
  </xsd:complexType>
  </xsd:element>
</xsd:schema>

I need to read the (second) XSD into C# and:

  1. check that it is a valid XSD, and
  2. validate documents against it.

Here is some C# to read in the schemata:

    XmlSchemaSet schemaSet = new XmlSchemaSet();
    foreach (string sd in Schemas)
    {
        using (XmlReader r = XmlReader.Create(new FileStream(sd, FileMode.Open)))
        {
            schemaSet.Add(XmlSchema.Read(r, null));
        }
    }
    schemaSet.CompilationSettings = new XmlSchemaCompilationSettings();
    schemaSet.Compile();

The .Compile() fails because "Type 'a:TheSimpleType' is not declared, or is not a simple type."

However, it works if either:

  • the namespace is removed from the schema, or
  • the namespace is added to the include.

The question is: how do I get C# to accept it without editing the schemata?

I suspect the problem is that although I have put both schemata into the XmlSchemaSet, I still need to tell C# that one is included into the other, i.e., it hasn't worked it out for itself. Indeed, if I only tell the XmlSchemaSet about the main XSD (and not the include) (both without (or with) namespaces) then "Type 'TheSimpleType' is not declared, or is not a simple type."

Thus this seems to be a question about resolving includes: how?!

Answer

Richard Barraclough picture Richard Barraclough · Apr 16, 2012

The problem is with the way the schema is opened for reading on the line:

XmlReader.Create(new FileStream(sd, FileMode.Open)

I had to write my own XmlResolver before I could see how the paths to the include files were being resolved: it was from the directory of the executable and not from the directory of the parent schema. The problem is that the parent schema was not getting its BaseURI set. Here's how the schema must be opened:

XmlReader.Create(new FileStream(pathname, FileMode.Open, FileAccess.Read),null, pathname)