Question

我正在连接到外部wsdl，并尝试来回发送一些消息。只需尝试发送一些数据并返回确认消息，就不需要任何复杂性。我现在的问题是我遇到了编组错误：

[com.sun.istack.internal.SAXException2: unable to marshal type "java.lang.String" as an element because it is missing an @XmlRootElement annotation]

这似乎来自这一行：

线程“主”中的异常javax.xml.bind.MarshalException -具有链接的例外： [com.sun.istack.internal.SAXException2：由于缺少@XmlRootElement批注，因此无法将类型“ java.lang.String”作为元素封送] 在com.sun.xml.internal.bind.v2.runtime.MarshallerImpl.write（MarshallerImpl.java:311）在com.sun.xml.internal.bind.v2.runtime.MarshallerImpl.marshal（MarshallerImpl.java:236）在javax.xml.bind.helpers.AbstractMarshallerImpl.marshal（AbstractMarshallerImpl.java:95）在DVSClient.main（DVSClient.java:59）由以下原因引起：com.sun.istack.internal.SAXException2：由于缺少@XmlRootElement批注，因此无法将类型“ java.lang.String”作为元素编组在com.sun.xml.internal.bind.v2.runtime.XMLSerializer.reportError（XMLSerializer.java:234）在com.sun.xml.internal.bind.v2.runtime.LeafBeanInfoImpl.serializeRoot（LeafBeanInfoImpl.java:127）在com.sun.xml.internal.bind.v2.runtime.property.SingleReferenceNodeProperty.serializeBody（SingleReferenceNodeProperty.java:100）在com.sun.xml.internal.bind.v2.runtime.ClassBeanInfoImpl.serializeBody（ClassBeanInfoImpl.java:345）在com.sun.xml.internal.bind.v2.runtime.ClassBeanInfoImpl.serializeBody（ClassBeanInfoImpl.java:336）在com.sun.xml.internal.bind.v2.runtime.ClassBeanInfoImpl.serializeBody（ClassBeanInfoImpl.java:336）在com.sun.xml.internal.bind.v2.runtime.XMLSerializer.childAsSoleContent（XMLSerializer.java:578）在com.sun.xml.internal.bind.v2.runtime.ClassBeanInfoImpl.serializeRoot（ClassBeanInfoImpl.java:326）在com.sun.xml.internal.bind.v2.runtime.XMLSerializer.childAsRoot（XMLSerializer.java:479）在com.sun.xml.internal.bind.v2.runtime.MarshallerImpl.write（MarshallerImpl.java:308）引用

我的主要代码如下：

public class Client {

public static void main(String[] args) throws MalformedURLException, ClassNotFoundException, IllegalAccessException, InstantiationException {


    URL UrlString = new URL("https://example.com.au:19443/Bus/exampleservice.svc?wsdl");
    String url02 = "https://example.com.au:19443/Bus/exampleservice.svc?wsdl";

    //Qname contains target namespace and name (port)
    QName qname = new QName("http://example/Contract/Service/Manager", "VerificationServiceBus", "Https");
    QName port = new QName("http://example/Contract/Service/Manager","DefaultSOAP11Binding_IVerification","Https");
    QName portName = new QName("Https");
    String serviceName = "VerificationServiceBus";
    CertificateRequest certificate = (new TestData()).test();
    URL newEndpoint = new URL("https://example.com.au:19443:19443/Bus/VerificationServiceBus.svc?");
    Service service = Service.create(UrlString, qname);

    try {

        IVerification myProxy = service.getPort(port,IVerification.class);

        VerificationResponse response = myProxy.verifyDocument(birthCertificate);
        System.out.println(response.getVerificationResultCode());

    }
    catch (IVerificationVerifyDocumentUnauthorizedAccessExceptionFaultFaultMessage ex) {
        System.out.println(ex.getMessage());

    }
}

}

我尝试手动进入某些类，并将@XMLRootElement添加到具有名称属性的类中，但仍存在相同的排序错误。

据我了解，编组和解组与将对象的内存表示形式转换为更适合转换的格式有关。虽然既然我所有的班级都生成了，不应该为我处理吗？

任何帮助将不胜感激。

谢谢

无法封送类型“ java.lang.String”作为元素，因为它缺少@XmlRootElement注释

0 个答案: