我必须使用2个肥皂调用网址才能在一个类中使用。 我的代码是:
public class Orderinfo extends Activity {
private static final String SOAP_ACTION = "http://xcart.com/data";
private static final String METHOD_NAME = "data";
private static final String NAMESPACE = "http://xcart.com";
private static final String URL = "http://192.168.1.168:8085/XcartLogin/services/RetailerWs?wsdl";
private static final String URL1 = "http://192.168.1.168:8085/XcartLogin/services/TodayC?wsdl";
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.table);
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.setOutputSoapObject(request);
HttpTransportSE ht = new HttpTransportSE(URL);
try {
ht.call(SOAP_ACTION, envelope);
SoapPrimitive response = (SoapPrimitive)envelope.getResponse();
SoapPrimitive s = response;
String str = s.toString();
String resultArr[] = str.split("&");//Result string will split & store in an array
TextView tv = (TextView) findViewById(R.id.textView44);
// TextView tv = new TextView(this);
for(int i = 0; i<resultArr.length;i++){
tv.append(resultArr[i]+"\n\n");
}
} catch (Exception e) {
e.printStackTrace();
}
HttpTransportSE ht1 = new HttpTransportSE(URL1);
try {
ht.call(SOAP_ACTION, envelope);
SoapPrimitive response = (SoapPrimitive)envelope.getResponse();
SoapPrimitive s = response;
String str = s.toString();
String resultArr[] = str.split("&");//Result string will split & store in an array
TextView tv = (TextView) findViewById(R.id.textView43);
// TextView tv = new TextView(this);
for(int i = 0; i<resultArr.length;i++){
tv.append(resultArr[i]+"\n\n");
}
} catch (Exception e) {
e.printStackTrace();
}
}
}
这里必须显示第一个soap呼叫输出是3.第二个soap呼叫输出是2.但是两个都只显示2个。为什么这里有什么例外。请帮助我。
答案 0 :(得分:0)
您在两次调用中都使用了一个信封请求,所谓的服务方法和参数是相同的