[英]JAVA web service receiving null parameters from Android Ksoap
我正在尝试从android使用Java Web服务。
到目前为止,这是我尝试过的:
private void CallWebServiceDummy() {
// TODO Auto-generated method stub
try {
if (android.os.Build.VERSION.SDK_INT > 9) {
StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder()
.permitAll().build();
StrictMode.setThreadPolicy(policy);
}
SoapSerializationEnvelope soapEnvelope = new SoapSerializationEnvelope(
SoapEnvelope.VER10);
soapEnvelope.dotNet = false;
SoapObject Request = new SoapObject(NAMESPACE, METHOD_NAME);
PropertyInfo pi = new PropertyInfo();
StringArraySerializer a = new StringArraySerializer();
a.add("hello"); a.add("world"); String n0 = NAMESPACE;
pi = new PropertyInfo(); pi.setName("a"); pi.setValue(a);
pi.setType(a.getClass()); pi.setNamespace(n0);
Request.addProperty(pi);
String b = "my name"; pi = new PropertyInfo(); pi.setName("b");
pi.setValue(b); Request.addProperty(pi);
soapEnvelope.setOutputSoapObject(Request);
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
androidHttpTransport.debug = true;
androidHttpTransport.call(SOAP_ACTION, soapEnvelope);
Log.d("test", "request: " + androidHttpTransport.requestDump);
Log.d("test", "response: " + androidHttpTransport.responseDump);
SoapObject resultsRequestSOAP = (SoapObject) soapEnvelope.bodyIn;
String c = resultsRequestSOAP.toString();
} catch (Exception e) {
Context context = getApplicationContext();
int duration = Toast.LENGTH_LONG;
Toast toast = Toast.makeText(context, e.getMessage(), duration);
toast.show();
toast.setGravity(Gravity.CENTER | Gravity.CENTER_HORIZONTAL, 0, 0);
}
}
我的Java Web服务代码:
package MyPackage;
public class WebServiceClass {
public String addnumbers(String[] a, String b) {
String c = new StringBuilder("This the String1 ").append(a[0]).append(" merged with String2 ").append(b).toString();
return c;
}
}
我的全局变量:
private static final String NAMESPACE = "http://MyPackage"; private
static final String URL =
"http://10.0.2.2:8080/WebService/services/WebServiceClass?wsdl"; private
static final String SOAP_ACTION = "urn:addnumbers"; private static final
String METHOD_NAME = "addnumbers";
问题:
我收到的回复是:
addnumbersResponse{return=This the String1 merged with String2 my name; }
第一个参数未发送到Web服务。 我试图删除此行:
soapEnvelope.dotNet = false; 但它仍然无法正常工作。
伙计们请帮帮我。 我被困了两天。 感谢您提供的任何帮助。
方法的第一个参数的类型是String数组,而不仅仅是String。 代替StringArraySerializer
,请尝试手动添加字符串。 您应该为此查看该页面: 向请求添加复杂对象数组 。
检查您的名称空间是否正确!
我花了三个小时解决相同的问题,然后才意识到WebService类中声明的名称空间与我创建SoapObject时使用的名称空间不同。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.