Hello, I've got a XML Schema and a XML instance of this schema. Now I want to generate a DataObject from the given XML instance. This works fine, but not as expected. If I've got a schema like this:
<?xml version="1.0" encoding="UTF-8"?> <xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" targetNamespace="http://www.organisation_01.com/xmlschema/v1.0/Person" xmlns:tns="http://www.organisation_01.com/xmlschema/v1.0/Person" elementFormDefault="qualified"> <xs:element name="person" type="personType"/> <xs:complexType name="personType"> <xs:all> <xs:element name="firstname" type="xs:string"/> <xs:element name="secondname" type="xs:string"/> </xs:all> </xs:complexType> </xs:schema> And my XML instance looks like this: <?xml version="1.0" encoding="UTF-8"?> <tns:person xmlns:tns="http://www.organisation_01.com/xmlschema/v1.0/Person" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.organisation_01.com/xmlschema/v1.0/Person http://localhost:8080/xmlschema/organisation_01/person.xsd"> <tns:firstname>Peter</tns:firstname> <tns:secondname>Pan</tns:secondname> </tns:person> I can read in the XMl instance with (scope is my HelperContext): InputStream in = SdoDroolsTest.class.getResourceAsStream("/person01.xml"); XMLDocument xml = scope.getXMLHelper().load(in); DataObject sdo_root = xml.getRootObject(); But the rootObject is not from type "personType" as expected. This is a big problem for me, because I've to do a rule based type mapping - this is quite difficult with generic types. My questions: 1. is there the possibility to declare sdo_root as "personType"... 2. or better - can Tuscany do this for me while generating the SDO from my XML? Thank's in advance for your help! Claus
