Question

我有使用java解析SOAP对象的程序。但是不可能返回解析的SOAP对象。以及如何在网址上传递它。我的节目是，

   public class MarshalDemo {
   public static void main(String[] args) throws Exception {
    Customer customer = new Customer();
    customer.id = 123;
    customer.firstName = "Jane";
    customer.lastName = "Doe";
    QName root = new QName("return");
    JAXBElement<Customer> je = new JAXBElement<Customer>(root, Customer.class, customer);

    XMLOutputFactory xof = XMLOutputFactory.newFactory();
    XMLStreamWriter xsw = xof.createXMLStreamWriter(System.out);
    xsw.writeStartDocument();
    xsw.writeStartElement("S", "Envelope", "http://schemas.xmlsoap.org/soap/envelope/");
    xsw.writeStartElement("S", "Body", "http://schemas.xmlsoap.org/soap/envelope/");
    xsw.writeStartElement("ns0", "findCustomerResponse", "http://service.jaxws.blog/");

    JAXBContext jc = JAXBContext.newInstance(Customer.class);
    Marshaller marshaller = jc.createMarshaller();
    marshaller.setProperty(Marshaller.JAXB_FRAGMENT, true);
    marshaller.marshal(je, xsw);

    xsw.writeEndDocument();
    xsw.close();
  }
 }

 class Customer {
   int id;
   String firstName;
   String lastName;
 }

Answer 1

您将希望利用JAX-WS来创建Web服务。您的服务类将如下所示。 JAX-WS将负责创建Envelope和Body元素以及通过网络发送数据。 JAX-WS利用JAXB作为对象到XML层，因此您只需要向Customer类添加任何必要的JAXB注释，JAX-WS会自动将它添加到消息中。

import javax.jws.*;

@WebService
public class FindCustomer {

 @WebMethod
 public Customer findCustomer(int id) {
  Customer customer = new Customer();
  customer.setId(id);
  customer.setFirstName("Jane");
  customer.setLastName("Doe");
  return customer;
 }

}

将转换后的SOAP / XML对象返回给url

1 个答案: