Android上的EditText电子邮件地址validation

我们怎样才能在android edittext上执行Email Validation ？ 我已经通过谷歌和SO，但我没有find一个简单的方法来validation它。

Answer 1

 public final static boolean isValidEmail(CharSequence target) { if (target == null) { return false; } else { return android.util.Patterns.EMAIL_ADDRESS.matcher(target).matches(); } } 

编辑::这将工作在Android 2.2以上！

Answer 2

要执行电子邮件validation，我们有很多方法，但最简单和最简单的方法是两种方法 。

1 –使用EditText(....).addTextChangedListener ，它保持在EditText box每个input触发，即email_id无效或有效

 /** * Email Validation ex:- [e-mail] */ 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._-]+@[az]+\\.+[az]+"; 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-使用if-else条件的最简单的方法。 使用getText（）获取EditText框string，并与为电子邮件提供的模式进行比较。 如果模式不匹配或macthes，onClick的button吐司消息。 它不会触发EditText框中每个字符的input。 简单的例子如下所示。

 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._-]+@[az]+\\.+[az]+"; // 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(); } 

Answer 3

使用此方法validation您的电子邮件格式。 将电子邮件作为string传递，如果格式正确则返回true，否则返回false。

 /** * validate your email address format. [e-mail] */ 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(); } 

Answer 4

这太简单了：

添加此方法来检查 电子邮件地址是否有效 ：

 private boolean isValidEmaillId(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string的EditText ：

 if(isValidEmaillId(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(); } 

完成

Answer 5

尝试这个：

 if (!emailRegistration.matches("[a-zA-Z0-9._-]+@[az]+.[az]+")) { editTextEmail.setError("Invalid Email Address"); } 

Answer 6

使用此方法validation电子邮件： –

  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

Answer 7

 public static boolean isEmailValid(String email) { boolean isValid = false; String expression = "^[\\w\\.-]+@([\\w\\-]+\\.)+[AZ]{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; } 

Answer 8

尝试这个

 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}" + ")+" ); 

并编辑文本

 final String emailText = email.getText().toString(); EMAIL_ADDRESS_PATTERN.matcher(emailText).matches() 

Answer 9

这是我为validation电子邮件地址而创build的示例方法，如果传递的string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\\-]+\\.)+[AZ]{2,4}$"; CharSequence inputStr = emailAddress; Pattern pattern = Pattern.compile(expression,Pattern.CASE_INSENSITIVE); Matcher matcher = pattern.matcher(inputStr); return matcher.matches(); }