Hi,
1. QVT -
Executing the following .qvto :
modeltype INTEROP uses "http://ms.com/interop";;
modeltype XSD uses "http://www.eclipse.org/xsd/2002/XSD";;
transformation interop2xsd(in interop: INTEROP , out XSD);
main(in interopModel: INTEROP::InteropSchema, out xsd: XSD::XSDSchema) {
xsd:= interopModel.map interop2Xsd();
}
mapping INTEROP::InteropSchema::interop2Xsd() : XSD::XSDSchema {
targetNamespace := 'DUMMY';
}
with OUT target file specifies as <something>.xsd, produces:
<?xml version="1.0" encoding="UTF-8"?>
<schema targetNamespace="DUMMY"/>
which does not look like a valid xsd schema root element.
2. Java-
On the other hand, running the following Java code:
class Schema {
private XSDFactory _xsdFactory;
private XSDSchema _schema;
private ByteArrayOutputStream _outputStream = new
ByteArrayOutputStream();
private Schema() {
_xsdFactory = XSDFactory.eINSTANCE;
initSchema();
}
private void initSchema() {
_schema = _xsdFactory.createXSDSchema();
_schema.setTargetNamespace("DUMMY");
_schema.setSchemaForSchemaQNamePrefix("xsd");
Map<String, String> qNamePrefixToNamespaceMap
=_schema.getQNamePrefixToNamespaceMap();
qNamePrefixToNamespaceMap.put("tns", _schema.getTargetNamespace());
qNamePrefixToNamespaceMap.put(_schema.getSchemaForSchemaQNamePrefix(),
XSDConstants.SCHEMA_FOR_SCHEMA_URI_2001);
_schema.updateElement();
}
private String serialize() {
XSDResourceImpl.serialize(_outputStream, _schema.getDocument());
return _outputStream.toString();
}
public static void main(String [ ] args) {
Schema schema = new Schema();
System.out.println(schema.serialize());
}
}
produces, in stdout:
<?xml version="1.0" encoding="UTF-8"?>
<xsd:schema xmlns:tns="DUMMY"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"; targetNamespace="DUMMY"/>
which looks like a valid xsd schema document root.
3. Question: how would one achieve the equivalence of 2. above,
with a QVT transform? I.e., how would one re-write the .qvto in 1. to
get the output produced in 2. ?
Thanks.
-Min