我在Apache Tomcat服务器中多次使用自顶向下方法(见下文)重新创建和部署了简单的Axis 2 Web服务。但是,我无法让这个简单的Web服务应用程序运行!
也就是说,当我提交简单的SOAP请求时(请参阅下面的“服务请求”),我收到以下错误:“[action]无法在接收方处理。”
问题:导致此错误的原因是什么:“”?我是否需要修改wsdl SoapAction值b4创建Web服务文件。请帮忙。
端点网址为:http://localhost:8080/NotifyEvt/services/EventNotifyService/
请求XML
<soap:Envelope
xmlns:soap="http://www.w3.org/2003/05/soap-envelope"
xmlns:even="http://veda-services/event-notify-v1-0.xsd"
xmlns:wsa="http://www.w3.org/2005/08/addressing">
<soap:Header>
<wsa:Action>http://localhost:8080/event/notifyEvent</wsa:Action>
</soap:Header>
<soap:Body>
<even:enquiry-id>23232323121266</even:enquiry-id>
</soap:Body>
</soap:Envelope>
错误回复
<soapenv:Body>
<soapenv:Fault>
<soapenv:Code>
<soapenv:Value>soapenv:Sender</soapenv:Value>
<soapenv:Subcode>
<soapenv:Value xmlns:wsa="http://www.w3.org/2005/08/addressing">wsa:ActionNotSupported</soapenv:Value>
</soapenv:Subcode>
</soapenv:Code>
<soapenv:Reason>
<soapenv:Text xml:lang="en-US">The [action] cannot be processed at the receiver.</soapenv:Text>
</soapenv:Reason>
<soapenv:Detail>
<wsa:ProblemAction xmlns:wsa="http://www.w3.org/2005/08/addressing">
<wsa:Action>{http://schemas.xmlsoap.org/wsdl/http/}//localhost:8080/event/notifyEventRequest</wsa:Action>
</wsa:ProblemAction>
请帮我解决问题。