Question

我正在创建一个非常简单的应用程序，以学习Parse功能。一路上我意识到我必须只使用用户名而不是电子邮件，（从存档的问题中得到这个，不确定现在是否有任何更改）。但在我的情况下，以下代码返回true，即使输入字段为空

protected void onCreate(Bundle savedInstanceState) {

        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login_activity);
        Parse.initialize(this, "#MASKED");

        emailLogin = (EditText)findViewById(R.id.loginEmail);
        passwordLogin = (EditText)findViewById(R.id.loginPassword);
        login_Login = (Button)findViewById(R.id.loginBtn);
        signup_Login = (Button)findViewById(R.id.loginSignup);

        signup_Login.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                Intent i = new Intent(getApplicationContext(),MainActivity.class);
                startActivity(i);
                finish();
            }
        });

        login_Login.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {

                String email = emailLogin.getText().toString().trim();
                String password = passwordLogin.getText().toString().trim();



                ParseUser.logInInBackground(email, password, new LogInCallback() {
                   @Override
                    public void done(ParseUser user, com.parse.ParseException e) {
                        if (e != null) {
                            // Hooray! The user is logged in.
                            Toast.makeText(Login_activity.this,"Sucessfully Logged in", Toast.LENGTH_LONG).show();
                            Intent i = new Intent(getApplicationContext(), HomePage.class);
                            startActivity(i);
                            finish();

                        } else {
                            // Signup failed. Look at the ParseException to see what happened.
                            Toast.makeText(Login_activity.this, e.getMessage(), Toast.LENGTH_LONG).show();
                            AlertDialog.Builder alertDiag = new AlertDialog.Builder(Login_activity.this);
                            alertDiag.setMessage(e.getMessage());
                            alertDiag.setTitle("Error");
                            alertDiag.setPositiveButton("Ok", new DialogInterface.OnClickListener() {
                                @Override
                                public void onClick(DialogInterface dialog, int which) {
                                    dialog.dismiss();
                                }
                            });

                            //AlertDialog dialog =
                                    alertDiag.create();
                            alertDiag.show();


                        }

                    }
                });





            }
        });
    }

Answer 1

   if (e != null) {
                        // Hooray! The user is logged in.

应该是：

if (e == null) {
                        // Hooray! The user is logged in.

因此，如果没有例外，则用户已成功登录。

此外，您的代码仅检查电子邮件和密码而非用户名。

使用电子邮件登录

1 个答案: