我有以下无法正常运行的SignUp活动:在用户收到验证电子邮件并尝试登录后,应用程序因NullPointerException崩溃,因为未创建Firebase实时数据库中的新用户条目。但是,我注意到,在调试期间,如果我在定义generateUser()
的位置设置断点,则会创建一个新的数据库条目(但应用程序崩溃的方式相同)。
这里有什么解决方案?
任何帮助都将受到高度赞赏。
更新:这里强调的不是NullPointerException,我可以处理。问题是为什么generateUser()
没有被调用。
public class SignUpActivity extends AppCompatActivity {
private EditText inputUsername, inputEmail, inputPassword;
private Button btnSignIn, btnSignUp;
private ProgressBar progressBar;
private FirebaseAuth auth;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
auth = FirebaseAuth.getInstance();
FirebaseUser user = auth.getCurrentUser();
if (user != null) {
if (user.isEmailVerified()) {
startActivity(new Intent(SignUpActivity.this, MainActivity.class));
finish();
}
}
setContentView(R.layout.activity_sign_up);
btnSignIn = findViewById(R.id.sign_in_button);
btnSignUp = findViewById(R.id.sign_up_button);
inputUsername = findViewById(R.id.username);
inputEmail = findViewById(R.id.email);
inputPassword = findViewById(R.id.password);
progressBar = findViewById(R.id.progressBar);
btnSignIn.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
startActivity(new Intent(SignUpActivity.this, SignInActivity.class));
}
});
btnSignUp.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
final String username = inputUsername.getText().toString().trim();
final String email = inputEmail.getText().toString().trim();
final String password = inputPassword.getText().toString().trim();
if (TextUtils.isEmpty(username)) {
Toast.makeText(getApplicationContext(), "Enter username!", Toast.LENGTH_SHORT).show();
return;
}
if (TextUtils.isEmpty(email)) {
Toast.makeText(getApplicationContext(), "Enter email address!", Toast.LENGTH_SHORT).show();
return;
}
if (TextUtils.isEmpty(password)) {
Toast.makeText(getApplicationContext(), "Enter password!", Toast.LENGTH_SHORT).show();
return;
}
if (password.length() < 6) {
Toast.makeText(getApplicationContext(), "Password too short, enter minimum 6 characters!", Toast.LENGTH_SHORT).show();
return;
}
progressBar.setVisibility(View.VISIBLE);
auth.createUserWithEmailAndPassword(email, password)
.addOnCompleteListener(SignUpActivity.this, new OnCompleteListener<AuthResult>() {
@Override
public void onComplete(@NonNull Task<AuthResult> task) {
progressBar.setVisibility(View.GONE);
if (!task.isSuccessful()) {
Toast.makeText(SignUpActivity.this, "Authentication failed.", Toast.LENGTH_SHORT).show();
} else {
new GenerateUserAsyncTask().execute(username, email, password, 0);
}
}
});
}
class GenerateUserAsyncTask extends AsyncTask<Object, Void, Void> {
@Override
protected Void doInBackground(Object... params) {
String username = (String) params[0];
String email = (String) params[1];
String password = (String) params[2];
int score = (int) params[3];
generateUser(username, email, password, score);
return null;
}
@Override
protected void onPostExecute(Void result) {
sendVerificationEmail();
}}
});
}
public void sendVerificationEmail() {
FirebaseUser user = auth.getCurrentUser();
if (user != null) {
user.sendEmailVerification()
.addOnCompleteListener(new OnCompleteListener<Void>() {
@Override
public void onComplete(@NonNull Task<Void> task) {
if (task.isSuccessful()) {
Toast.makeText(SignUpActivity.this, "Signup successful, verification email sent", Toast.LENGTH_SHORT).show();
auth.signOut();
startActivity(new Intent(SignUpActivity.this, SignInActivity.class));
finish();
} else {
Toast.makeText(SignUpActivity.this, "Failed to send email!", Toast.LENGTH_SHORT).show();
}
progressBar.setVisibility(View.GONE);
}
});
}
}
public void generateUser(String username, String email, String password, int score) {
FirebaseDatabase database = Utils.getDatabase();
DatabaseReference users = database.getReference("users");
User user = new User(username, email, password, score);
users.child(auth.getUid()).setValue(user);
}
}
答案 0 :(得分:0)
我找到了解决方案。该问题是由实时数据库安全规则引起的:它们仅允许用户在经过身份验证时写入数据库。但是,在我的代码中,调用generateUser()
时,用户尚未完全通过身份验证。所以我需要在用户点击验证邮件中的链接后在数据库中生成新条目。