scalaxb and xsd:choice

Question

Using http://scalaxb.org/

  <xsd:complexType name="Address">
    <xsd:choice>
      <xsd:element ref="ExternalAddress" />
      <xsd:element ref="InternalAddress" />
    </xsd:choice>
  </xsd:complexType>


val internalAddrress = InternalAddress(...);  // this works.
val address : Address = internalAddrress;     // error: type mismatch

how do I need to modify this code to make it work?

There is some info here, and I played with DataRecord and others, but I didn't make it work. http://scalaxb.org/narrower-choice

Answer 1

The question was incorrectly formulated. Here the correct question and answer:

 <xsd:complexType name="Address">
    <xsd:choice>
      <xsd:element name="externalAddress" type="ExternalAddress" />
      <xsd:element name="internalAddress" type="InternalAddress" />
    </xsd:choice>
  </xsd:complexType>


val internalAddress = InternalAddress(...);  // this works.
val address = Address(scalaxb.DataRecord(None, Some("internalAddress"), internalAddress));  // now this works.

See https://github.com/eed3si9n/scalaxb/issues/138#issuecomment-3943088

Answer 2

What exactly are you expressing with <xs:choice>? xs:choice declares child element of the complex type that could either be ExternalAddress or InternalAddress. But judging from your Scala code, you might be trying to express an address type that could either be external or internal.

In that case, complex type extension is the way to go. See http://scalaxb.org/running-scalaxb for an example of USAddress extending Address. This generates Addressable trait which are supertype for both Address and USAddress.