我的 .net 网络服务显然正在运行 soap 1.2(通过检查 .wsdl)并且我一直在尝试访问 helloworld 网络服务进行测试，但我遇到了错误。 顺便说一下，我正在尝试通过模拟器来做到这一点。

所以当我使用 soap 1.2 版本时，我收到错误消息“无法处理没有有效操作参数的请求。请提供有效的 soap” 我想知道我缺少什么以及我应该做什么。

谢谢!

我已经做过的事情:

• 添加android使用互联网的权限
• Soap 版本 1.1 和 1.2 的变化
• 从 SoapObject 更改为 Object(对于 soap 1.1 和 1.2)
• 模拟器使用 10.0.2.2
• 检查地址和方法名称中的拼写错误

我的代码:

 private static final String NAMESPACE = "http://localhost/WebService/";
 private static final String URL = "http://10.0.2.2:1672/Eventurous/WsEventurousMobile.asmx";
 private static final String HelloWorld_SOAP_ACTION = "http://localhost/WebService/HelloWorld";
 private static final String METHOD_NAME1 = "HelloWorld";



...
...

public static String GetHelloWorld() {

  SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME1);
  SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(
     SoapEnvelope.VER12);
  envelope.dotNet = true;
  envelope.setOutputSoapObject(request);
  HttpTransportSE androidHttpTransport = new HttpTransportSE(URL,60000);

  try {
   androidHttpTransport.setXmlVersionTag("<?xml version=\"1.0\" encoding=\"utf-8\"?>");
    androidHttpTransport.call(HelloWorld_SOAP_ACTION, envelope);

   SoapObject response = (SoapObject)envelope.getResponse(); 
   String result = response.getProperty(0).toString(); 

   return result;
   } catch (Exception e) {
   return e.toString();
  }

 }

Soap 1.2 版错误

Code: soap:Sender, Reason: System.Web.Services.Protocols.SoapException: Unable to handle request without a valid action parameter. Please supply a valid soap action.

 at System.Web.Services.Protocols.Soap12ServerProtocolHelper.RouteRequest()

at System.Web.Services.Protocols.SoapServerProtocol.RouteRequest(SoapServerMessage message)

at System.Web.Services.Protocols.SoapServerProtocol.Initialize()

at System.Web.Services.Protocols.ServerProtocolFactory.Create(Type type, HttpContext context, HttpRequest request, HttpResponse response, Boolean

Soap 1.1 版错误

SoapFault - faultcode: 'soap:Client' faultstring: 'System.Web.Services.Protocols.SoapException: Server did not recognize the value of HTTP Header SOAPAction: http://localhost/WebService/HelloWorld.

 at System.Web.Services.Protocols.Soap11ServerProtocolHelper.RouteRequest()

at System.Web.Services.Protocols.SoapServerProtocol.RouteRequest(SoapServerMessage message)

at System.Web.Services.Protocols.SoapServerProtocol.Initialize()

at System.Web.Services.Protocols.ServerProtocolFactory.Create(Type type, HttpContext context, HttpRequest request, HttpResponse response, Boolean& abortProcessing)' faultactor: 'null' detail: org.kxml2.kdom.Node@413c9098

最佳答案

使用 SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);

代替

SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(
     SoapEnvelope.VER12);

并删除此行 androidHttpTransport.setXmlVersionTag("<?xml version=\"1.0\" encoding=\"utf-8\"?>");

而不是 SoapObject response = (SoapObject)envelope.getResponse();

使用SoapObject response = (SoapObject)envelope.bodyIn;

它会帮助你。如果仍然出现错误，请写信给我。

关于android - soap 1.2 android 请提供有效的 soap 操作，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10910359/