完整的答案如下：

Question

我正在发送SOAP请求，服务器抱怨SOAPAction标头为空。我想我做对了，但显然我不是。 Wireshark显示它没有设置。

@Test
public void testLogin() throws Exception {
    StringBuffer loginXml = new StringBuffer();
    loginXml.append("<soapenv:Envelope xmlns:soapenv=\"http://www.w3.org/2003/05/soap-envelope\" xmlns:ns=\"http://example.com/xyz/2010/08\">");
    loginXml.append("  <soapenv:Header>");
    loginXml.append("    <ns:loginOperationDetails>");
    loginXml.append("    </ns:loginOperationDetails>");
    loginXml.append("  </soapenv:Header>");
    loginXml.append("  <soapenv:Body>");
    loginXml.append("    <ns:LogIn>");
    loginXml.append("      <ns:logInInfo>");
    loginXml.append("        <ns:CustomerAccountId>customer1</ns:CustomerAccountId>");
    loginXml.append("        <ns:Username>JDoe</ns:Username>");
    loginXml.append("        <ns:Password>abc123</ns:Password>");
    loginXml.append("      </ns:logInInfo>");
    loginXml.append("    </ns:LogIn>");
    loginXml.append("  </soapenv:Body>");
    loginXml.append("</soapenv:Envelope>");

    WebServiceTemplate webServiceTemplate = new WebServiceTemplate();
    MessageFactory msgFactory = MessageFactory.newInstance(SOAPConstants.SOAP_1_2_PROTOCOL);
    SaajSoapMessageFactory newSoapMessageFactory = new SaajSoapMessageFactory(msgFactory);
    webServiceTemplate.setMessageFactory(newSoapMessageFactory);

    String uri = "http://xyz.example.com/xyz_1.0/membership.svc/ws";
    webServiceTemplate.setDefaultUri(uri);

    StreamSource source = new StreamSource(new StringReader(loginXml.toString()));
    StreamResult result = new StreamResult(System.out);

    boolean resultReturned = false;
    try {
        resultReturned = webServiceTemplate.sendSourceAndReceiveToResult(source, 
            new SoapActionCallback("http://example.com/xyz/2010/08/MembershipService/LogIn"), 
            result);
    } 
    catch (SoapFaultClientException sfe) {
        logger.error("SoapFaultClientException resultReturned: " + resultReturned, sfe);
        fail();
    }
}

我从服务器回来的错误说：

500 Internal Server Error
The SOAP action specified on the message, '', does not match the HTTP SOAP Action, 'http://example.com/xyz/2010/08/MembershipService/LogIn'.

Answer 1

完整的答案如下：

当您使用WebServiceTemplate作为类与Web服务进行通信时，我不明白为什么，但它没有正确填充HTTP标头。

一些WSDL有一个部分说：

<soap:operation
            soapAction="SOMELINK"
            style="document" />

而WebServiceTemplate忽略了这一部分。上述错误表示标头中的soapAction参数为空。它应该不是。查看Wireshark。我做了 - 使用一些Chrome Soap客户端和Spring。第二个标题无效。

要解决此问题，您需要按照此处的第6.2.4节进行操作：http://docs.spring.io/spring-ws/sites/2.0/reference/html/client.html

它所说的基本上是使用WebServiceMessageCallback接口自行添加标题部分。您可以在参考文献中阅读更多内容。

基本上它最终会像这样：

 webServiceTemplate.marshalSendAndReceive(o, new WebServiceMessageCallback() {

    public void doWithMessage(WebServiceMessage message) {
        ((SoapMessage)message).setSoapAction("http://tempuri.org/Action");
    }
});

您可以在哪里正确设置标头值。也为我工作。整天阅读。

Answer 2

我解决了这个问题，但从未发布过答案。以下是我最终得到的结果：

public WebServiceTemplate getWebServiceTemplate() throws SOAPException {
  if (webServiceTemplate == null) {
    final MessageFactory msgFactory = MessageFactory.newInstance(SOAPConstants.SOAP_1_2_PROTOCOL);
    final SaajSoapMessageFactory newSoapMessageFactory = new SaajSoapMessageFactory(msgFactory);
    webServiceTemplate = new WebServiceTemplate(newSoapMessageFactory);
  }   

  return webServiceTemplate;
}

public Object sendReceive(Object requestObject, ArrayList<String> classesToMarshall, final String action)
        throws ClassNotFoundException, SoapFaultException, SoapFaultClientException, WebServiceTransportException,
        IllegalStateException, SOAPException {

  final WebServiceTemplate wst = getWebServiceTemplate();

    final SoapMarshallUtil smu = getSoapMarshallUtil();
    smu.configureMarshaller(wst, classesToMarshall);

    // soap 1.2
    SoapActionCallback requestCallback = new SoapActionCallback(action) {
        public void doWithMessage(WebServiceMessage message) {
            SaajSoapMessage soapMessage = (SaajSoapMessage) message;
            SoapHeader soapHeader = soapMessage.getSoapHeader();

            QName wsaToQName = new QName("http://www.w3.org/2005/08/addressing", "To", "wsa");
            SoapHeaderElement wsaTo =  soapHeader.addHeaderElement(wsaToQName);
            wsaTo.setText(uri);

            QName wsaActionQName = new QName("http://www.w3.org/2005/08/addressing", "Action", "wsa");
            SoapHeaderElement wsaAction =  soapHeader.addHeaderElement(wsaActionQName);
            wsaAction.setText(action);
        }
    };

    Object responseObject = wst.marshalSendAndReceive(this.uri, requestObject, requestCallback);
    return responseObject;
}

Answer 3

另一种解决方法是添加 interceptor 并在handleRequest()方法中设置soapAction，以使入站接收MessageContext，{派生;

之后，您可以轻松设置调用SoapMessage方法的soapAction。

这是拦截器类的代码：

setSoapAction()

，当然还要将拦截器添加到public class SecurityInterceptor implements ClientInterceptor { @Override public boolean handleRequest(MessageContext messageContext) throws WebServiceClientException { SoapMessage soapMessage = (SoapMessage) messageContext.getRequest(); soapMessage.setSoapAction("mySoapAction"); return true; } //TODO:: other methods and constructor.. }：

WebTemplate

Spring-WS客户端没有设置SOAPAction头

3 个答案:

完整的答案如下：