227

我们怎样才能Email Validation在edittextin上表演android？我已经通过 google & SO 但我没有找到一种简单的方法来验证它。

4

9 回答 9

751

爪哇：

public static boolean isValidEmail(CharSequence target) {
    return (!TextUtils.isEmpty(target) && Patterns.EMAIL_ADDRESS.matcher(target).matches());
}

科特林：

fun CharSequence?.isValidEmail() = !isNullOrEmpty() && Patterns.EMAIL_ADDRESS.matcher(this).matches()

编辑：它将适用于Android 2.2+ ！

编辑：添加缺失;

于 2013-04-04T09:58:32.720 回答
82

要执行电子邮件验证，我们有很多方法，但简单和最简单的方法是两种方法。

1-使用EditText(....).addTextChangedListenerwhich 不断触发EditText boxie email_id 中的每个输入无效或有效

/**
 * Email Validation ex:- tech@end.com
*/


final EditText emailValidate = (EditText)findViewById(R.id.textMessage); 

final TextView textView = (TextView)findViewById(R.id.text); 

String email = emailValidate.getText().toString().trim();

String emailPattern = "[a-zA-Z0-9._-]+@[a-z]+\\.+[a-z]+";

emailValidate .addTextChangedListener(new TextWatcher() { 
    public void afterTextChanged(Editable s) { 

    if (email.matches(emailPattern) && s.length() > 0)
        { 
            Toast.makeText(getApplicationContext(),"valid email address",Toast.LENGTH_SHORT).show();
            // or
            textView.setText("valid email");
        }
        else
        {
             Toast.makeText(getApplicationContext(),"Invalid email address",Toast.LENGTH_SHORT).show();
            //or
            textView.setText("invalid email");
        }
    } 
    public void beforeTextChanged(CharSequence s, int start, int count, int after) {
    // other stuffs 
    } 
    public void onTextChanged(CharSequence s, int start, int before, int count) {
    // other stuffs 
    } 
});

2-if-else使用条件的 最简单方法。使用 getText() 获取 EditText 框字符串并与为电子邮件提供的模式进行比较。如果模式不匹配或不匹配，按钮的 onClick 会显示一条消息。它不会在 EditText 框中的每个字符输入时触发。如下所示的简单示例。

final EditText emailValidate = (EditText)findViewById(R.id.textMessage); 

final TextView textView = (TextView)findViewById(R.id.text); 

String email = emailValidate.getText().toString().trim();

String emailPattern = "[a-zA-Z0-9._-]+@[a-z]+\\.+[a-z]+";

// onClick of button perform this simplest code.
if (email.matches(emailPattern))
{
Toast.makeText(getApplicationContext(),"valid email address",Toast.LENGTH_SHORT).show();
}
else 
{
Toast.makeText(getApplicationContext(),"Invalid email address", Toast.LENGTH_SHORT).show();
}
于 2012-10-18T05:20:26.347 回答
44

我是这样做的：

添加此方法以检查 电子邮件地址是否有效：

private boolean isValidEmailId(String email){

    return Pattern.compile("^(([\\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})$").matcher(email).matches();
     }

现在检查EditText的字符串：

if(isValidEmailId(edtEmailId.getText().toString().trim())){
  Toast.makeText(getApplicationContext(), "Valid Email Address.", Toast.LENGTH_SHORT).show();
}else{       
  Toast.makeText(getApplicationContext(), "InValid Email Address.", Toast.LENGTH_SHORT).show();
}

完毕

于 2013-02-02T05:14:21.657 回答
39

使用此方法验证您的电子邮件格式。将 email 作为字符串传递，如果格式正确则返回 true，否则返回 false。

/**
 * validate your email address format. Ex-akhi@mani.com
 */
public boolean emailValidator(String email) 
{
    Pattern pattern;
    Matcher matcher;
    final String EMAIL_PATTERN = "^[_A-Za-z0-9-]+(\\.[_A-Za-z0-9-]+)*@[A-Za-z0-9]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$";
    pattern = Pattern.compile(EMAIL_PATTERN);
    matcher = pattern.matcher(email);
    return matcher.matches();
}
于 2012-10-18T05:23:39.670 回答
12

试试这个：

if (!emailRegistration.matches("[a-zA-Z0-9._-]+@[a-z]+\.[a-z]+")) {
 editTextEmail.setError("Invalid Email Address");
}
于 2012-10-18T05:27:44.087 回答
9

使用此方法验证 EMAIL：- 

 public static boolean isEditTextContainEmail(EditText argEditText) {

            try {
                Pattern pattern = Pattern.compile("^[_A-Za-z0-9-]+(\\.[_A-Za-z0-9-]+)*@[A-Za-z0-9]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$");
                Matcher matcher = pattern.matcher(argEditText.getText());
                return matcher.matches();
            } catch (Exception e) {
                e.printStackTrace();
                return false;
            }
        }

如果您有任何疑问，请告诉我？

于 2013-09-25T10:15:33.603 回答
8 
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;
}
于 2012-10-18T05:30:07.707 回答
6

试试这个

public static final Pattern EMAIL_ADDRESS_PATTERN = Pattern.compile(

              "[a-zA-Z0-9\\+\\.\\_\\%\\-\\+]{1,256}" +
              "\\@" +
              "[a-zA-Z0-9][a-zA-Z0-9\\-]{0,64}" +
              "(" +
              "\\." +
              "[a-zA-Z0-9][a-zA-Z0-9\\-]{0,25}" +
              ")+"
          );

并在 tne 中编辑文本

final String emailText = email.getText().toString();
EMAIL_ADDRESS_PATTERN.matcher(emailText).matches()
于 2012-10-18T05:28:35.490 回答
6

这是我创建的用于验证电子邮件地址的示例方法，如果传递的字符串参数是有效的电子邮件地址，则返回 true，否则返回 false。

private boolean validateEmailAddress(String emailAddress){
    String  expression="^[\\w\\-]([\\.\\w])+[\\w]+@([\\w\\-]+\\.)+[A-Z]{2,4}$";  
       CharSequence inputStr = emailAddress;  
       Pattern pattern = Pattern.compile(expression,Pattern.CASE_INSENSITIVE);  
       Matcher matcher = pattern.matcher(inputStr);  
       return matcher.matches();
}
于 2012-10-19T07:33:27.807 回答