Question

我正在尝试从Android项目调用web服务，但我无法传递任何参数。

我需要建立一个这样的请求：

<?xml version="1.0" encoding="UTF-8"?>
<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns1="www.mywebservice.com">
  <SOAP-ENV:Body>
    <ns1:MyMethodName>
      <ns1:appUser>0</ns1:appUser>
    </ns1:MyMethodName>
  </SOAP-ENV:Body>
</SOAP-ENV:Envelope>

但是使用ksoap2，我得到这样的结果：

<v:Envelope xmlns:i="http://www.w3.org/2001/XMLSchema-instance" xmlns:d="http://www.w3.org/2001/XMLSchema" xmlns:c="http://schemas.xmlsoap.org/soap/encoding/" xmlns:v="http://schemas.xmlsoap.org/soap/envelope/">
  <v:Header />
  <v:Body>
    <MyMethodName xmlns="www.mywebservice.com">
      <appUser>0</appUser>
    </MyMethodName>
  </v:Body>
</v:Envelope>

我的代码是：

SoapObject request = new SoapObject(NAMESPACE,METHOD_NAME);
request.addProperty("appUser",0);

SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet = true;
envelope.setAddAdornments(false);
envelope.implicitTypes = true;

envelope.setOutputSoapObject(request);

HttpTransportSE httpTransport = new HttpTransportSE(URL);
httpTransport.debug = true;

httpTransport.call(SOAP_ACTION, envelope);

尝试了很多示例，但似乎没有任何工作 - 我无法构建正确的XML请求。

Answer 1

我解决了这个问题：

创建了一个新类，它扩展了 SoapSerializationEnvelope 并覆盖了它的方法 write - 我已经修改了它以满足我的需求：

public class MyNewClass extends SoapSerializationEnvelope {

        public MyNewClass (int version) {
            super(version);
        }

        @Override
        public void write(XmlSerializer writer) throws IOException {
            env = "http://schemas.xmlsoap.org/soap/envelope/";
            String tem = "www.mywebservice.com";
            writer.startDocument("UTF-8", true);
            writer.setPrefix("SOAP-ENV", env);
            writer.setPrefix("ns1", tem);
            writer.startTag(env, "Envelope");
            writer.startTag(env, "Body");
            writer.startTag(tem, "MyMethodName");
            writeBody(writer);
            writer.endTag(tem, "MyMethodName");
            writer.endTag(env, "Body");
            writer.endTag(env, "Envelope");
            writer.endDocument();
        }
    }

而不是使用 SoapSerializationEnvelope ：

SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);

我使用我新创建的类：

MyNewClass envelope = new MyNewClass (SoapEnvelope.VER11);

如何使用android-ksoap2构建正确的XML请求？

1 个答案: