SOAP12 缺少肥皂动作 - 弹簧

Question

I am trying to make a simple SOAP request but I am getting this error:

org.springframework.ws.soap.client.SoapFaultClientException: Unable to handle request without a valid action parameter. Please supply a valid soap action.

I am using soap12 . Here is my code:

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

Jaxb2Marshaller marshaller = new Jaxb2Marshaller();
marshaller.setPackagesToScan("packageName");
marshaller.afterPropertiesSet();

webServiceTemplate.setMarshaller(marshaller);
webServiceTemplate.afterPropertiesSet();

Object response = webServiceTemplate.marshalSendAndReceive("https://www.w3schools.com/xml/tempconvert.asmx", temperature);

I tried setting Content-Type with WebServiceMessageCallback to text/xml but it didn't get overriden.

Answer 1

I found a solution by adding:

MessageFactory messageFactory = MessageFactory.newInstance(SOAPConstants.SOAP_1_2_PROTOCOL);
SaajSoapMessageFactory saajSoapMessageFactory = new SaajSoapMessageFactory(messageFactory);
saajSoapMessageFactory.setSoapVersion(SoapVersion.SOAP_12);
saajSoapMessageFactory.afterPropertiesSet();