单击特定按钮时关闭应用程序

Question

当我点击登录按钮时应用程序保持关闭我认为错误来自代码但我不知道错误到底在哪里

任何帮助！

只是这个按钮将我从他的应用程序中踢出来，其他按钮效果很好！

public class login extends AppCompatActivity {


    EditText username;
    EditText password;
    Button userlogin;
    FirebaseAuth firebseAuth;


    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);


        username = findViewById(R.id.login_username);
        password = findViewById(R.id.login_password);
        userlogin = findViewById(R.id.login_btn);

        firebseAuth = FirebaseAuth.getInstance();
        userlogin.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                firebseAuth.signInWithEmailAndPassword(username.getText().toString(),password.getText().toString())
                        .addOnCompleteListener(new OnCompleteListener<AuthResult>() {
                            @Override
                            public void onComplete(@NonNull Task<AuthResult> task) {
                                if(task.isSuccessful()){
                                    startActivity(new Intent(login.this,profile.class));
                                }else{
                                    Toast.makeText(login.this,"hellow", Toast.LENGTH_LONG).show();
                                }

                            }
                        });
            }
        });

    }
}`

Answer 1

您可以检查用户名和密码字符串是否为空，然后发出请求：

String passwordStr = password.getText().toString() 
String usernameStr = username.getText().toString() 
if (!TextUtils.isEmpty(passwordStr) && !TextUtils.isEmpty(usernameStr) {
    firebseAuth.signInWithEmailAndPassword(usernameStr, passwordStr).addOnCompleteListener(....)
}

如果字符串为空，firebseAuth.signInWithEmailAndPassword() 似乎会抛出异常。