View activities when a IF statement is true

Question

Need help with switching to an activity if my form details match the string. So when the login button is pressed it will display main activity. This is the code I have so far:

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.login);
    setupVariables();
}

private void setupVariables() {
    username = (EditText) findViewById(R.id.username);
    password = (EditText) findViewById(R.id.password);
    login = (Button) findViewById(R.id.button);

}

public void Login(View view) {
    if (username.getText().toString().equals("admin") && password.getText().toString().equals("secret")) {
        /* switches activity if the username and password matches the string */
    }
    else {

    }
}

}

Possible duplicate of [How to start new activity on button click](http://stackoverflow.com/questions/4186021/how-to-start-new-activity-on-button-click) — George Mulligan, Apr 02 '16 at 16:38
@joydroider Just did, error Lenght.Short) can't resolve symbol. — DeeMoMo, Apr 02 '16 at 17:12
@DeeMoMo Sorry mistaken i have edited my answer. Please check edited part. — Jay Rathod, Apr 02 '16 at 17:22

score 0 · Answer 1 · edited Apr 02 '16 at 16:37

0

Supposing the Login as the onClick function of the login button, you can use Intent like following:

Intent intent = new Intent(CurrentActivity.this, nextActivity.class);
startActivity(intent);

You are good to go!

edited Apr 02 '16 at 16:37

Amir_P

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answered Apr 02 '16 at 16:33

score 0 · Answer 2 · answered Apr 02 '16 at 16:36

you will it like this

    login.setOnClickListener(new View.OnclickListener(){
     @override
     public void onClick(View v){
      Login();
    }
    });
public void Login() {
    if (username.getText().toString().equals("admin") && password.getText().toString().equals("secret")) {
     Intent intent = new Intent(this,NewActivity.class);
startActivity(intent);
    }
    else {  
    Toast.makeText(this,"Cant't Login",Lenght.Short).show();

    }

P Sharma · Answer 3 · 2016-04-02T17:15:15.503

0

use this onclick method

login.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                 if (username.getText().toString().equals("admin")&&password.getText().toString().equals("secret")) {
                Intent intent = new Intent(Login.this, MainActivity.class);
                startActivity(intent);
                }
               else {  
                Toast.makeText(getApplicationContext(), "Wrong username or password", 
 Toast.LENGTH_LONG).show();
               }
            }
        });

I am assuming that Login.java is your current class EDIT--> try this

login.setOnClickListener(new OnClickListener(){
    @Override
    //On click function
    public void onClick(View view) {
if (username.getText().toString().equals("admin")&&password.getText().toString().equals("secret"))
{    
        //Create the intent to start another activity
        Intent intent = new Intent(Login.this, MainActivity.class);
        startActivity(intent);
    }
else {
Toast.makeText(this, "Wrong username or password", Toast.LENGTH_SHORT).show();       
}
});

edited Apr 02 '16 at 17:15

answered Apr 02 '16 at 16:38

P Sharma

184
9

I am receiving errors on can't resolve symbol for setOnClickListener, the v in (View v) and LENGTH_LONG @Praval Sharma – DeeMoMo Apr 02 '16 at 16:49
I edited my answer and paste another code for your errors @DeeMoMo – P Sharma Apr 02 '16 at 17:02
Again, now have an can't resolve symbol for 'button' at the beginning. The (new OnClickListener() has an error. And both (View view) and (view.getContext() have errors. – DeeMoMo Apr 02 '16 at 17:05
you have to use 'login' instead of 'button' because your 'login' is your button object and replace view.getcontext() with 'Login.this'. Here I am assuming Login is your current class. – P Sharma Apr 02 '16 at 17:14
Also i have changed code for your convenience @DeeMoMo – P Sharma Apr 02 '16 at 17:16

score 0 · Answer 4 · answered Apr 02 '16 at 16:46

For this please use "Intent" class, used to move from one activity to another activity or service.

Intent intent=new Intent(first_activity.class, Second_activity.class);
startactivity(intent);

or you can use

Intent intent=new Intent(getapplicationcontext(),Second_activity.class);
startactivity(intent)

Jay Rathod · Accepted Answer · 2016-04-02T17:21:40.140

For Switching and working with activity.

You first might read details for this. Follow this.

http://developer.android.com/training/basics/firstapp/starting-activity.html

And try this with your code.

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.login);
    setupVariables();

}

private void setupVariables() {
    username = (EditText) findViewById(R.id.username);
    password = (EditText) findViewById(R.id.password);
    login = (Button) findViewById(R.id.button);
    login.setOnClickListener(new View.OnClickListener() {
       @Override
       public void onClick(View v) {
           Login();
       }
    });

}

public void Login() {
    if (username.getText().toString().equals("admin") && password.getText().toString().equals("secret")) {
        // Here you can switch your activity.
        Intent i = new Intent(getApplicationContext(), SecondScreen.class);
        StartActivity(i);
    }
    else {
        Toast.makeText(this,"Invalid credentials.",Toast.LENGTH_SHORT).show();
    }
}

And you not need to pass view parameter as you need to call it directly on Button Click Listener method.

Edit

Toast.makeText(this,"Invalid credentials.",Toast.LENGTH_SHORT).show();

Hope this will help you to understand basics.

@DeeMoMo Glad to Help :). Please upvote it also if u can Thanks :). — Jay Rathod, Apr 02 '16 at 17:24

View activities when a IF statement is true

5 Answers5