如何在android中對電子郵件進行驗證？

Question

我正在開發一個應用程序，其中我有一個電子郵件編輯文本字段，我在其中進行了一些驗證，但沒有正確輸入電子郵件，例如jj@gm.cim但除了這是錯誤的右邊驗證是jj @ gmail.com。

我該怎麼做？

這裏是代碼： -

public boolean validation(String mobile,String pass,String email){ 
    if (mobile!=null && mobile.length() >= 7 && mobile.length() <= 15) { 
     if (pass.length() >= 4 && pass.length() <= 8) { 
      if (isValidEmail(email)) { 
       return true; 
      } else { 
       m_EmailEditText.setError("Invalid EmailID"); 
       return false; 
      } 
     } else { 
      m_PasswordEditText.setError("password must be between 4 to 8 characters long"); 
      return false; 
     } 
    } else { 
     m_MobileEditText.setError("mobile number must be between 7 to 51 characters long"); 
     return false; 
    } 
} 
// This is validation for email……….and link to above code…. 
private boolean isValidEmail(String email) { 
    String emailPattern = "^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*@" 
      + "[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$"; 

    Pattern pattern = Pattern.compile(emailPattern); 
    Matcher matcher = pattern.matcher(email); 
    return matcher.matches(); 
} 

Answer 1

如果你想使用正則表達式，然後嘗試這種方式來獲得有效的電子郵件

public static boolean isEmailValid(String email) { 
    boolean isValid = false; 

    String expression = "^[\\w\\.-]+@([\\w\\-]+\\.)+[A-Z]{2,4}$"; 
    CharSequence inputStr = email; 

    Pattern pattern = Pattern.compile(expression, Pattern.CASE_INSENSITIVE); 
    Matcher matcher = pattern.matcher(inputStr); 
    if (matcher.matches()) { 
     isValid = true; 
    } 
    return isValid; 
} 

OR 如果你想使用默認的驗證嘗試這種方式

boolean isEmailValid(CharSequence email) { 
    return android.Util.Patterns.EMAIL_ADDRESS.matcher(email).matches(); 
} 

Answer 2

您也可以使用，

public class EmailValidator { 

private Pattern pattern; 
private Matcher matcher; 

private static final String EMAIL_PATTERN = 
    "^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*@" 
    + "[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$"; 

public EmailValidator() { 
    pattern = Pattern.compile(EMAIL_PATTERN); 
} 

/** 
* Validate hex with regular expression 
* 
* @param hex 
*   hex for validation 
* @return true valid hex, false invalid hex 
*/ 
public boolean validate(final String hex) { 

    matcher = pattern.matcher(hex); 
    return matcher.matches(); 

} 
} 

Answer 3

功能用於確認EMAILID

private boolean isEmailValid(String email) { 

    String regExpn = "^(([\\w-]+\\.)+[\\w-]+|([a-zA-Z]{1}|[\\w-]{2,}))@" + "((([0-1]?[0-9]{1,2}|25[0-5]|2[0-4][0-9])\\.([0-1]?" + "[0-9]{1,2}|25[0-5]|2[0-4][0-9])\\." 
      + "([0-1]?[0-9]{1,2}|25[0-5]|2[0-4][0-9])\\.([0-1]?" + "[0-9]{1,2}|25[0-5]|2[0-4][0-9])){1}|" + "([a-zA-Z]+[\\w-]+\\.)+[a-zA-Z]{2,4})$"; 
    CharSequence inputStr = email; 
    Pattern pattern = Pattern.compile(regExpn, Pattern.CASE_INSENSITIVE); 
    Matcher matcher = pattern.matcher(inputStr); 
    if (matcher.matches()) 
     return true; 
    else 
     return false; 

} 

Answer 4

參見機器人模式

它`機器人inbuild正則表達式的方法。 https://stackoverflow.com/a/7882950/3879847