Database 错误的密码/电子邮件错误身份验证活动-使用SQLite
Onclick()方法有什么问题?或verifyFromSqlite()?尝试使用刚刚提供给注册表的数据登录,为什么按下登录按钮的输出只是一个错误,即密码/电子邮件错误Database 错误的密码/电子邮件错误身份验证活动-使用SQLite,database,sqlite,authentication,android-sqlite,Database,Sqlite,Authentication,Android Sqlite,Onclick()方法有什么问题?或verifyFromSqlite()?尝试使用刚刚提供给注册表的数据登录,为什么按下登录按钮的输出只是一个错误,即密码/电子邮件错误 public class LoginActivity extends AppCompatActivity implements View.OnClickListener{ private final AppCompatActivity activity = LoginActivity.this; private Neste
public class LoginActivity extends AppCompatActivity implements View.OnClickListener{
private final AppCompatActivity activity = LoginActivity.this;
private NestedScrollView nestedScrollView;
private TextInputLayout textInputLayoutEmail;
private TextInputLayout textInputLayoutPassword;
private TextInputEditText textInputEditTextEmail;
private TextInputEditText textInputEditTextPassword;
private AppCompatButton appCompatButtonLogin;
private AppCompatTextView textViewLinkRegister;
private InputValidation inputValidation;
private DatabaseHelper databaseHelper;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_login);
getSupportActionBar().hide();
initViews();
initListeners();
initObjects();
}
private void initViews(){
nestedScrollView = (NestedScrollView) findViewById(R.id.nestedScrollView);
textInputLayoutEmail= (TextInputLayout) findViewById(R.id.textInputLayoutEmail);
textInputLayoutPassword= (TextInputLayout) findViewById(R.id.textInputLayoutPassword);
textInputEditTextEmail=(TextInputEditText) findViewById(R.id.textInputEditTextEmail);
textInputEditTextPassword=(TextInputEditText) findViewById(R.id.textInputEditTextPassword);
appCompatButtonLogin = (AppCompatButton) findViewById(R.id.appCompatButtonLogin);
textViewLinkRegister= (AppCompatTextView) findViewById(R.id.textViewLinkRegister);
}
private void initListeners(){
appCompatButtonLogin.setOnClickListener(this);
textViewLinkRegister.setOnClickListener(this);
}
private void initObjects(){
databaseHelper = new DatabaseHelper(activity);
inputValidation = new InputValidation(activity);
}
@Override
public void onClick(View v){
switch (v.getId()){
case R.id.appCompatButtonLogin:
verifyFromSQLite();
break;
case R.id.textViewLinkRegister:
Intent intentRegister = new Intent(getApplicationContext(), RegisterActivity.class);
startActivity(intentRegister);
break;
}
}
private void verifyFromSQLite(){
if (!inputValidation.isInputEditTextFilled(textInputEditTextEmail, textInputLayoutEmail, getString(R.string.error_message_email))){
return;
}
if (!inputValidation.isInputEditTextEmail(textInputEditTextEmail, textInputLayoutEmail, getString(R.string.error_message_email))){
return;
}
if (!inputValidation.isInputEditTextFilled(textInputEditTextPassword, textInputLayoutPassword, getString(R.string.error_message_password))){
return;
}
if(databaseHelper.checkUser(textInputEditTextEmail.getText().toString().trim()
, textInputEditTextPassword.getText().toString().trim())){
Intent accountsIntent = new Intent(activity, UsersActivity.class);
accountsIntent.putExtra("EMAIL", textInputEditTextEmail.getText().toString().trim());
emptyInputEditText();
startActivity(accountsIntent);
} else {
Snackbar.make(nestedScrollView, getString(R.string.error_valid_email_password), Snackbar.LENGTH_LONG).show();
}
}
private void emptyInputEditText(){
textInputEditTextEmail.setText(null);
textInputEditTextPassword.setText(null);
}
}
我相信这会做错事
if(databaseHelper.checkUser(textInputEditTextEmail.getText().toString().trim()
, textInputEditTextPassword.getText().toString().trim())){
Intent accountsIntent = new Intent(activity, UsersActivity.class);
accountsIntent.putExtra("EMAIL", textInputEditTextEmail.getText().toString().trim());
emptyInputEditText();
startActivity(accountsIntent);
} else {
Snackbar.make(nestedScrollView, getString(R.string.error_valid_email_password), Snackbar.LENGTH_LONG).show();
}
}
请让我知道我做错了什么,因为我已经尝试了所有其他答案,但它不会点击我:)为什么注册并尝试登录后会向我发送错误消息 假设checkUser方法与以前的方法相同,即:-
public boolean checkUser(String password, String email){
String[] columns = {
COLUMN_USER_ID
};
SQLiteDatabase db= this.getWritableDatabase();
String selection = COLUMN_USER_EMAIL + " = ? " + "AND "+ COLUMN_USER_PASSWORD+" =? ";
String[] selectionArgs = { email,password };
Cursor cursor = db.query(TABLE_USER,
columns,
selection,
selectionArgs,
null,
null,
null);
int cursorCount = cursor.getCount();
cursor.close();
db.close();
if(cursorCount > 0){
return true;
}
return false;
}
然后,您将电子邮件作为密码传递,并将密码作为电子邮件传递。尝试更改:-
if(databaseHelper.checkUser(textInputEditTextEmail.getText().toString().trim()
, textInputEditTextPassword.getText().toString().trim())){
Intent accountsIntent = new Intent(activity, UsersActivity.class);
accountsIntent.putExtra("EMAIL", textInputEditTextEmail.getText().toString().trim());
emptyInputEditText();
startActivity(accountsIntent);
} else {
Snackbar.make(nestedScrollView, getString(R.string.error_valid_email_password), Snackbar.LENGTH_LONG).show();
}
}
致:-
不,先生,这行不通,我想你发布了完全相同的代码来更改。有什么变化吗,因为我看不见?相同的电子邮件或密码有误。。它不会保存我的注册数据。@LaviniaRotaru区别在于调用
checkUser
方法的第一个参数是密码(textinputtextpassword.getText().toString().trim()
)而不是电子邮件(textinputtextemail.getText().toString().trim()
)是与密码相对的第二个参数,因为checkUser
方法的签名是password,email而不是email,password。
if(databaseHelper.checkUser( textInputEditTextPassword.getText().toString().trim()
,textInputEditTextEmail.getText().toString().trim())){
Intent accountsIntent = new Intent(activity, UsersActivity.class);
accountsIntent.putExtra("EMAIL", textInputEditTextEmail.getText().toString().trim());
emptyInputEditText();
startActivity(accountsIntent);
} else {
Snackbar.make(nestedScrollView, getString(R.string.error_valid_email_password), Snackbar.LENGTH_LONG).show();
}
}