如何驗證從EditText上電子郵件和密碼輸入註冊用戶

-1

之前

如何從EditText上註冊使用Android Studio用戶之前驗證電子郵件&密碼輸入...如何驗證從EditText上電子郵件和密碼輸入註冊用戶

電子郵件：必須包含@符號和其他一般要求密碼：必須> 6位數字。

請修改此代碼。我的代碼是...

public class RegisterActivity extends AppCompatActivity implements View.OnClickListener{ 

private Button buttonRegister; 
private EditText editTextEmail; 
private EditText editTextPassword; 
private TextView textViewSignin; 

private ProgressDialog progressDialog; 

private FirebaseAuth firebaseAuth; 

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

    firebaseAuth=FirebaseAuth.getInstance(); 
    if (firebaseAuth.getCurrentUser()!=null){ 
     //profile activity here 
     finish(); 
     startActivity(new Intent(getApplicationContext(),ProfileActivity.class)); 
    } 

    progressDialog = new ProgressDialog(this); 

    buttonRegister = (Button)findViewById(R.id.buttonRegister); 
    editTextEmail = (EditText)findViewById(R.id.editTextEmail); 
    editTextPassword = (EditText)findViewById(R.id.editTextPassword); 
    textViewSignin = (TextView)findViewById(R.id.textViewSignin); 

    buttonRegister.setOnClickListener(this); 
    textViewSignin.setOnClickListener(this); 
} 

private void registerUser(){ 
    final String email = editTextEmail.getText().toString().trim(); 
    String password = editTextPassword.getText().toString().trim(); 

    //uset cannot go any further without entering password 
    if(TextUtils.isEmpty(email)){ 
     //email is empty 
     Toast.makeText(this, "Please enter email",Toast.LENGTH_SHORT).show(); 
     //stopping execution further 
     return; 
    } 
    if(TextUtils.isEmpty(password)){ 
     //password is empty 
     Toast.makeText(this, "Please enter password",Toast.LENGTH_SHORT).show(); 
     //stopping execution further 
     return; 
    } 

    //if validations are ok 
    //we will first show progressbar 
    progressDialog.setMessage("Registerating User..."); 
    progressDialog.show(); 

    firebaseAuth.createUserWithEmailAndPassword(email, password) 
      .addOnCompleteListener(this, new OnCompleteListener<AuthResult>() { 
       @Override 
       public void onComplete(@NonNull Task<AuthResult> task) { 

        if(task.isSuccessful()){ 
         //user is successfully registered. we will start profile activity here 
         FirebaseUser user = firebaseAuth.getCurrentUser(); 
         Toast.makeText(RegisterActivity.this, "Authentication success. " + user.getUid(), Toast.LENGTH_SHORT).show(); 
         progressDialog.hide(); 
         finish(); 
         startActivity(new Intent(getApplicationContext(),ProfileActivity.class)); 
        } 
        else{ 
         Toast.makeText(RegisterActivity.this, "Could not register. please try again", Toast.LENGTH_SHORT).show(); 
         progressDialog.hide(); 
        } 

       } 
      }); 
} 

@Override 
public void onClick(View v) { 
    if (v==buttonRegister){ 
     registerUser(); 
    } 
    if(v==textViewSignin){ 
     finish(); 
     startActivity(new Intent(this, LoginActivity.class)); 
    } 
} 
}

來源

2017-05-20 Atif Pervaiz

，添加的inputType屬性的EditText和密碼，編程獲取它，使用 'gettext的（）。的toString（）'，然後檢查長度這個字符串。 –

試試這個我的朋友

String emailAddress = etSignInEmail.getText().toString().trim(); 
    if (etSignInPassword.getText().toString().length() < 6) { 
    etSignInPassword.setError(getString("password minimum contain 6 character)); 
    etSignInPassword.requestFocus(); 
    } 
    if (etSignInPassword.getText().toString().equals("")) { 
    etSignInPassword.setError(getString("please enter password)); 
    etSignInPassword.requestFocus(); 
    } 
    if (!android.util.Patterns.EMAIL_ADDRESS.matcher(emailAddress).matches()) { 
    etSignInEmail.setError(getString("please enter valid email address)); 
    etSignInEmail.requestFocus(); 
    } 
    if (etSignInEmail.getText().toString().equals("")) { 
    etSignInEmail.setError(getString("please enter email address)); 
    etSignInEmail.requestFocus(); 
    } 
    if (!emailAddress.equals("") && 
    etSignInPassword.getText().toString().length() >= 6 && 
    !etSignInPassword.getText().toString().trim().equals("") 
    && android.util.Patterns.EMAIL_ADDRESS.matcher(emailAddress).matches()) { 
    // do your action 
    }

來源

2017-05-20 09:34:43

感謝弟弟我根據你的邏輯編輯了我的代碼，現在它的工作很完美... –

你爲什麼不試試這個。

android.util.Patterns.EMAIL_ADDRES

或

public static boolean EMailValidation(String emailstring) { 
     if (null == emailstring || emailstring.length() == 0) { 
      return false; 
     } 
     Pattern emailPattern = Pattern 
       .compile("^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*@" 
         + "[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$"); 
     Matcher emailMatcher = emailPattern.matcher(emailstring); 
     return emailMatcher.matches(); 
    }

在XML

來源

2017-05-20 10:54:02

如何驗證從EditText上電子郵件和密碼輸入註冊用戶

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