我有一个安卓应用程序。在这里,我必须使用soap webservices创建带有mysql连接的登录页面。
我的代码在 android 2.2 版本上运行良好,但在 android 4.0 上运行不正常。
所以在这里我必须使用异步任务。但我不知道异步任务。请帮我解决这些问题。
编辑:
我使用以下代码:
public class AndroidLoginExampleActivity extends Activity {
private final String NAMESPACE = "http://ws.userlogin.com";
private final String URL = "http://111.223.128.10:8085/AndroidLogin/services/Login?wsdl";
private final String SOAP_ACTION = "http://ws.userlogin.com/authentication";
private final String METHOD_NAME = "authentication";
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
Button login = (Button) findViewById(R.id.btn_login);
login.setOnClickListener(new View.OnClickListener() {
public void onClick(View arg0) {
new LongOperation().execute();
}
} );
}
class LongOperation extends AsyncTask<String, Void, String> {
private ProgressDialog Dialog = new ProgressDialog(LoginActivity.this);
@Override
protected String doInBackground(String... aurl) {
// TODO Auto-generated method stub
loginAction();
return null;
}
protected void onPreExecute() {
Dialog.setMessage("Loading...");
Dialog.show();
}
protected void onPostExecute(String resultGot) {
Dialog.dismiss();
}
}
private void loginAction(){
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
EditText userName = (EditText) findViewById(R.id.tf_userName);
String user_Name = userName.getText().toString();
EditText userPassword = (EditText) findViewById(R.id.tf_password);
String user_Password = userPassword.getText().toString();
//Pass value for userName variable of the web service
PropertyInfo unameProp =new PropertyInfo();
unameProp.setName("userName");//Define the variable name in the web service method
unameProp.setValue(user_Name);//set value for userName variable
unameProp.setType(String.class);//Define the type of the variable
request.addProperty(unameProp);//Pass properties to the variable
//Pass value for Password variable of the web service
PropertyInfo passwordProp =new PropertyInfo();
passwordProp.setName("password");
passwordProp.setValue(user_Password);
passwordProp.setType(String.class);
request.addProperty(passwordProp);
envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.setOutputSoapObject(request);
androidHttpTransport = new HttpTransportSE(URL);
runOnUiThread(new Runnable() {
@Override
public void run() {
try{
androidHttpTransport.call(SOAP_ACTION, envelope);
SoapPrimitive response = (SoapPrimitive)envelope.getResponse();
TextView result = (TextView) findViewById(R.id.tv_status);
result.setText(response.toString());
}
catch(Exception e){
}
});
}
}
现在也遇到同样的错误:
在那之前:
现在我必须在目标版本 android 4.0 上创建应用程序,这意味着它在 android 2.2 上运行良好。但它在 android 4.0 版本上不起作用。请检查我的代码并给我解决方案。