我必须开发一个Android应用程序。
在这里，我必须获取beloning用户的电子邮件，并在android textview.how上显示该电子邮件，我可以开发这些。请给我任何想法。
这是我的网络服务代码：
public class Login {
public String authentication(String username,String password){
String retrievedUserName = "";
String retrievedPassword = "";
String retrievedEmail = "";
String status = "";
try{
Class.forName("com.mysql.jdbc.Driver");
Connection con = DriverManager.getConnection("jdbc:mysql://localhost:3306/xcart-432pro","root","");
PreparedStatement statement = con.prepareStatement("SELECT * FROM xcart_customers WHERE login = '"+username+"'");
ResultSet result = statement.executeQuery();
while(result.next()){
retrievedUserName = result.getString("login");
retrievedPassword = result.getString("password");
retrievedEmail = result.getString("email");
}
if(retrievedUserName.equals(username)&&retrievedPassword.equals(password)&&!(retrievedUserName.equals("") && retrievedPassword.equals(""))){
status = retrievedEmail;
}
else {
status = "Login fail!!!";
}
}
catch(Exception e){
e.printStackTrace();
}
return status;
}
}
在这里我必须运行这些web服务代码意味着如果登录成功意味着已成功显示电子邮件ID。如果我的登录不正确意味着显示登录失败消息。那么我的webservice代码是正确的。
如果我的登录成功，则意味着必须在android textview上显示电子邮件。否则必须显示登录失败的消息。
但在这里我遇到了一个问题：
如果我输入正确的登录详细信息意味着时间也显示登录失败的消息。但是如果我的登录成功意味着显示电子邮件，则需要o / p，否则登录失败消息。
我的代码出了什么问题。请为我提供解决方案。
这是我的android端代码：
public class CustomerLogin extends Activity {
private static final String SPF_NAME = "vidslogin";
private static final String USERNAME = "login";
private static final String PASSWORD = "password";
private static final String PREFS_NAME = null;
private String login;
String mGrandTotal,total,mTitle,mTotal,mQty,mCost;
EditText username,userPassword;
private final String NAMESPACE = "http://xcart.com";
private final String URL = "http://10.0.0.75:8085/XcartLogin/services/Login?wsdl";
private final String SOAP_ACTION = "http://xcart.com/authentication";
private final String METHOD_NAME = "authentication";
private String uName;
/**Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.customer_login);
username = (EditText) findViewById(R.id.tf_userName);
userPassword = (EditText) findViewById(R.id.tf_password);
Button login = (Button) findViewById(R.id.btn_login);
login.setOnClickListener(new View.OnClickListener() {
public void onClick(View arg0) {
loginAction();
}
});
}
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);
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.setOutputSoapObject(request);
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);
try{
androidHttpTransport.call(SOAP_ACTION, envelope);
SoapPrimitive response = (SoapPrimitive)envelope.getResponse();
String status = response.toString();
TextView email = (TextView) findViewById(R.id.tv_status);
email.setText(response.toString());
if(status.equals(email))
{
if(isUserValidated && isPasswordValidated)
{
Intent intent = new Intent(CustomerLogin.this,PayPalIntegrationActivity.class);
startActivity(intent);
}
}
else
{
LayoutInflater inflater = getLayoutInflater();
View layout = inflater.inflate(R.layout.toast_custom_layout,
(ViewGroup) findViewById(R.id.toast_layout_root));
Toast toast = new Toast(getApplicationContext());
toast.setGravity(Gravity.TOP, 0, 30);
toast.setDuration(Toast.LENGTH_LONG);
toast.setView(layout);
toast.show();
}
}
catch(Exception e){
}
}
}
修改
我在这里编写了像手段
这样的代码 if(status.equals("krishnaveniveeman@gmail.com"))
在输入正确的登录deatils后，在textview中收到电子邮件。
在这里，我如何从webservice获取电子邮件，并在textview上设置电子邮件。
我必须编写像
这样的代码 if(status.equals(email))
表示仅显示登录失败消息。但我的登录详细信息仅为正确。
请有人给我这些解决方案。