Android ksoap2参数问题

Question

我试图将参数传递给我的服务，代码运行但服务从未收到参数？调用有效，我只需添加变量然后将其取回，当取回它时，我发现网络服务从未收到它！

感谢您的帮助

    final String SOAP_ACTION = "http://NathofGod.com/GetCategoryById";
    final String METHOD_NAME = "GetCategoryById";
    final String NAMESPACE = " http://NathofGod.com/";
    final String URL = "http://10.0.2.2:4021/Service1.asmx";

SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);

 PropertyInfo pi1 = new PropertyInfo();
        pi1.setName("name");
        pi1.setValue("the name");
        pi1.setType(String.class);
        pi1.setNamespace(NAMESPACE);
        request.addProperty(pi1);

 SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
        envelope.dotNet = true;
        envelope.setOutputSoapObject(request);

       HttpTransportSE conn = new HttpTransportSE(URL);

        try
        {
            conn.call(SOAP_ACTION, envelope);
            SoapObject response = (SoapObject)envelope.getResponse();
        }

        catch(Exception e)
        {
            e.printStackTrace();
        }

Answer 1

不确定为什么不起作用，但我记得使用它

request.addProperty("name", "my_Name");

并且工作正常，否则你可能想检查服务器端......

Answer 2

这行代码是我的问题！！！

envelope.dotNet = true;

删除它

Answer 3

尝试按照说明进行调试在维基上。

Answer 4

我稍微清理了一下代码并将其放入一个函数中。我不确定是否有什么不同，但这段代码可以工作。感谢您的回复。

public SoapObjectsoap() 抛出 IOException、XmlPullParserException
{

    SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME); 
    request.addProperty("name", "myname"); 
    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11); 
    envelope.dotNet = true;
    envelope.setOutputSoapObject(request); 
    HttpTransportSE conn = new HttpTransportSE(URL);

    conn.call(SOAP_ACTION, envelope); //send request
    SoapObject result=(SoapObject)envelope.getResponse(); 
    return result;
 }

Answer 5

最终字符串 NAMESPACE = " http://NathofGod.com/";

更改为

final String NAMESPACE = "http://NathofGod.com/";

删除空白并按照请求 XML 上的方式创建命名空间。请注意，它也区分大小写。