我正在使用 Resteasy 开发 RestWebService。基本实现工作正常。知道我试图通过休息返回一个复杂的对象......实际上它很容易......我想。由于我的嵌套对象(地址),我遇到了问题......
我尝试的是这样的:
@XmlRootElement(name = "person")
@XmlAccessorType(XmlAccessType.FIELD)
public class Person implements Serializable {
private static final long serialVersionUID = 1199647317278849602L;
private String uri;
private String vName;
private String nName;
private Address address;
.....
@XmlElementWrapper(name="Former-User-Ids")
@XmlElement(name="Adress")
public Address getAddress() {
return address;
}
....
地址如下所示:
@XmlRootElement(name = "address")
@XmlAccessorType(XmlAccessType.FIELD)
public class Address {
private String uri;
private String street;
private String city;
public String getCity() {
return city;
}
public String getStreet() {
return street;
}
.... Restservice 看起来像这样。没有地址对象,它工作得很好。
@Path("/getPersonXML/{personNumber}")
@GET
@Produces(MediaType.APPLICATION_XML)
public Patient getPatientXML(@PathParam("personNumber") String personNumber) throws ParseException {
Address a1 = new Address("de.person/address/" + "432432","Teststret12","TestCity", "32433", "TestCountry", "081511833");
Patient p1 = new Person();
p1.setAddress(a1);
p1.setUri("de.spironto/person/"+ "432432");
p1.setnName("Power");
p1.setvName("Max");
return p1;
}
目前我总是得到一个
javax.xml.bind.JAXBException:
有任何想法吗?