Question

@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);

           Button  btnLogin=(Button)this.findViewById(R.id.Button01);
         btnLogin.setOnClickListener(new OnClickListener() {

             @Override
             public void onClick(View v) {
                 if((txtUserName.getText().toString()).equals(txtPassword.getText().toString())){
                       Toast.makeText(LoginappActivity.this, "Login Successful",Toast.LENGTH_LONG).show();
                 } 
                 else
                 {
                       Toast.makeText(LoginappActivity.this, "Invalid Login",Toast.LENGTH_LONG).show(); 
                 }
               }
             }
         );

           Button next = (Button) findViewById(R.id.Button01);
           next.setOnClickListener(new View.OnClickListener() {   

                public void onClick(View view) {
                Intent myIntent = new Intent(view.getContext(), AddName.class);
                startActivityForResult(myIntent, 0); 
            }});    
       }    
 }

当应用程序在模拟器中运行时，它已成功编译，但我不知道是否成功登录或登录无效。

Answer 1

您需要将此代码放在Login Button的onClickListener中。修改后的代码应为

Button  btnLogin=(Button)this.findViewById(R.id.Button01);
             btnLogin.setOnClickListener(new OnClickListener() {

                 @Override
                 public void onClick(View v) {
                     // TODO Auto-generated method stub
if((txtUserName.getText().toString()).equals(txtPassword.getText().toString())){
            Toast.makeText(LoginappActivity.this, "Login Successful",Toast.LENGTH_LONG).show();
           } else{
            Toast.makeText(LoginappActivity.this, "Invalid Login",Toast.LENGTH_LONG).show(); 
            }
              }
                 }
             });

Answer 2

在按钮单击事件中写入If ... Else。

 Button  btnLogin=(Button)this.findViewById(R.id.Button01);
         btnLogin.setOnClickListener(new OnClickListener() {

             @Override
             public void onClick(View v) {
                 if((txtUserName.getText().toString()).equals(txtPassword.getText().toString())){
                       Toast.makeText(LoginappActivity.this, "Login Successful",Toast.LENGTH_LONG).show();
                 } 
                 else
                 {
                       Toast.makeText(LoginappActivity.this, "Invalid Login",Toast.LENGTH_LONG).show(); 
                 }
               }
             }
         });

我无法看到应用程序是否成功登录

2 个答案: