Question

当我输入用户名和密码（如果两者都相同）时，我创建了一个登录页面，它用于登录，但用户名和密码留空，当我点击模拟器上的登录按钮时，它显示登录成功当用户名和密码保留为balnk时，它不应该登录

 @Override
         public void onCreate(Bundle savedInstanceState) {
         super.onCreate(savedInstanceState);
         setContentView(R.layout.main);

     txtUserName=(EditText)this.findViewById(R.id.txtUname);
     txtPassword=(EditText)this.findViewById(R.id.txtPwd);
     btnLogin=(Button)this.findViewById(R.id.btnLogin);
     btnLogin.setOnClickListener(new OnClickListener() {

             @Override
             public void onClick(View v) {            

                    if((txtUserName.getText().toString()).equals(txtPassword.getText().toString())){
                    Intent myIntent = new Intent(v.getContext(), SaveData.class);
                    startActivityForResult(myIntent, 1);
                    Toast.makeText(LoginappActivity.this, "Login Successful",Toast.LENGTH_LONG).show();
                    } else{
                    Toast.makeText(LoginappActivity.this, "Invalid Login",Toast.LENGTH_LONG).show(); 
                    }
               }
           });
 }           
       }

Answer 1

因为空白用户名和密码也相同。您需要检查用户是否未输入任何值，然后提示它。更合适的是，使用TextUtils.isEmpty来完成这项工作。

public void onClick(View v) { 
   String username = txtUserName.getText().toString();
   String password = txtPassword.getText().toString();

   if(!TextUtils.isEmpty(username) && username.equals(password)){

          //do your stuff
   }
}

Answer 2

首先尝试一些验证，确认任何一个字段是否为空您验证的代码是

 btnLogin.setOnClickListener(new OnClickListener() {

                 @Override
                 public void onClick(View v) {            

                        if(txtUserName.getText().toString()).equals("")){
                             //Show Message
                        }else if(txtPassword.getText().toString().equals("")){
                             //Show Message
                        }else if(!(txtUserName.getText().toString()).equals(txtPassword.getText().toString())){
                           Toast.makeText(LoginappActivity.this, "Invalid Login",Toast.LENGTH_LONG).show(); 
                        }else{
                            Intent myIntent = new Intent(v.getContext(), SaveData.class);
                            startActivityForResult(myIntent, 1);
                            Toast.makeText(LoginappActivity.this, "Login Successful",Toast.LENGTH_LONG).show();
                        }
                   }
               });

Answer 3

你只是检查两者是否相等，两者都是空白的，所以如果条件为真......

所以你应该检查两个字段是否为空......

<强>喜欢...

String strName = txtUserName.getText().toString();
String strPass = txtPwd.getText().toString();


     if(strName!=null&&strName.trim().length!=0&&strPass!=null && strPass.trim().length!=0){

    Intent myIntent = new Intent(v.getContext(), SaveData.class);
    startActivityForResult(myIntent, 1);
    Toast.makeText(LoginappActivity.this, "Login Successful",Toast.LENGTH_LONG).show();

  }else{
                Toast.makeText(LoginappActivity.this, "input your name and password",Toast.LENGTH_LONG).show();
           }

Answer 4

只需在txtUserName.getText().toString().length() > 0

中添加条件&& operator和if

if((txtUserName.getText().toString().length() > 0) && (txtUserName.getText().toString()).equals(txtPassword.getText().toString())){
       Intent myIntent = new Intent(v.getContext(), SaveData.class);
       startActivityForResult(myIntent, 1);
       Toast.makeText(LoginappActivity.this, "Login Successful",Toast.LENGTH_LONG).show();
       } else{
          Toast.makeText(LoginappActivity.this, "Invalid Login",Toast.LENGTH_LONG).show(); 
          }
     }

Answer 5

这是因为""等于""（空字符串）

您可能想要添加此

[...]
if(txtUserName.getText().toString().length() > 0 && (txtUserName.getText().toString()).equals(txtPassword.getText().toString()))
[...]

没有输入用户名和密码我的应用程序正在记录

5 个答案: