大家好我有一个edittext，我想在我的Editttext中编写电子邮件验证 这是一个xml代码
<EditText
android:id="@+id/mail"
android:layout_width="match_parent"
android:layout_height="48dp"
android:layout_alignLeft="@+id/phone"
android:layout_below="@+id/phone"
android:layout_marginRight="33dp"
android:layout_marginTop="10dp"
android:background="@drawable/edit_background"
android:ems="10"
android:hint="E-Mail"
android:inputType="textEmailAddress"
android:paddingLeft="20dp"
android:textColor="#7e7e7e"
android:textColorHint="#7e7e7e" >
</EditText>
这是一个java代码
emailInput = mail.getText().toString().trim();
emailPattern = "^[_A-Za-z0-9-]+(\\.[_A-Za-z0-9-]+)*@[A-Za-z0-9]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$";
if (emailInput.matches(emailPattern)) {
Toast.makeText(getActivity(), "valid email address",
Toast.LENGTH_SHORT).show();
} else {
Toast.makeText(getActivity(), "Invalid email address",
Toast.LENGTH_SHORT).show();
mail.setBackgroundResource(R.drawable.edit_red_line);
}
我无法验证。吐司信息始终是“无效的电子邮件地址” 我究竟做错了什么？如果有人知道解决方案，请帮助我
答案 0 :(得分：89)
为什么不使用：
public final static boolean isValidEmail(CharSequence target) {
return !TextUtils.isEmpty(target) && android.util.Patterns.EMAIL_ADDRESS.matcher(target).matches();
}
根据建议here。
答案 1 :(得分：2)
我发布了非常简单易用的电子邮件验证答案，而不使用任何字符串模式。
1.在按钮上设置点击监听器....
button_resetPassword.setOnClickListener(new View.OnClickListener()
{
@Override
public void onClick(View v)
{
CharSequence temp_emilID=username.getText().toString();//here username is the your edittext object...
if(!isValidEmail(temp_emilID))
{
username.requestFocus();
username.setError("Enter Correct Mail_ID ..!!");
or
Toast.makeText(getApplicationContext(), "Enter Correct Mail_ID", Toast.LENGTH_SHORT).show();
}
else
{
correctMail..
//Your action...
}
});
2.拨打isValidEmail（），即..
public final static boolean isValidEmail(CharSequence target)
{
if (TextUtils.isEmpty(target))
{
return false;
} else {
return android.util.Patterns.EMAIL_ADDRESS.matcher(target).matches();
}
}
我希望它对你有帮助......
答案 2 :(得分：1)
Android电子邮件验证最简单的方法
String validemail= "[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}" +
")+";
String emal=email.getText().toString();
Matcher matcherObj = Pattern.compile(validemail).matcher(emal);
if (matcherObj.matches()) {
Toast.makeText(getApplicationContext(), "enter
all details", Toast.LENGTH_SHORT).show();
}
else
{
Toast.makeText(getApplicationContext(),"please enter
valid email",Toast.LENGTH_SHORT).show();
}
答案 3 :(得分：1)
请尝试以下代码：
scoreWk1这很好用。
答案 4 :(得分：0)
% 10 == 0答案 5 :(得分：0)
这是登录验证的完整代码......
public class LoginActivity extends AppCompatActivity {
private static final String TAG = "LoginActivity";
private static final int REQUEST_SIGNUP = 0;
@Bind(R.id.input_email) EditText _emailText;
@Bind(R.id.input_password) EditText _passwordText;
@Bind(R.id.btn_login) Button _loginButton;
@Bind(R.id.link_signup) TextView _signupLink;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_login);
ButterKnife.bind(this);
_loginButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
login();
}
});
_signupLink.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
// Start the Signup activity
Intent intent = new Intent(getApplicationContext(), SignupActivity.class);
startActivityForResult(intent, REQUEST_SIGNUP);
finish();
overridePendingTransition(R.anim.push_left_in, R.anim.push_left_out);
}
});
}
public void login() {
Log.d(TAG, "Login");
if (!validate()) {
onLoginFailed();
return;
}
_loginButton.setEnabled(false);
final ProgressDialog progressDialog = new ProgressDialog(LoginActivity.this,
R.style.AppTheme_Dark_Dialog);
progressDialog.setIndeterminate(true);
progressDialog.setMessage("Authenticating...");
progressDialog.show();
String email = _emailText.getText().toString();
String password = _passwordText.getText().toString();
// TODO: Implement your own authentication logic here.
new android.os.Handler().postDelayed(
new Runnable() {
public void run() {
// On complete call either onLoginSuccess or onLoginFailed
onLoginSuccess();
// onLoginFailed();
progressDialog.dismiss();
}
}, 3000);
}
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
if (requestCode == REQUEST_SIGNUP) {
if (resultCode == RESULT_OK) {
// TODO: Implement successful signup logic here
// By default we just finish the Activity and log them in automatically
this.finish();
}
}
}
@Override
public void onBackPressed() {
// Disable going back to the MainActivity
moveTaskToBack(true);
}
public void onLoginSuccess() {
_loginButton.setEnabled(true);
finish();
}
public void onLoginFailed() {
Toast.makeText(getBaseContext(), "Login failed", Toast.LENGTH_LONG).show();
_loginButton.setEnabled(true);
}
public boolean validate() {
boolean valid = true;
String email = _emailText.getText().toString();
String password = _passwordText.getText().toString();
if (email.isEmpty() || !android.util.Patterns.EMAIL_ADDRESS.matcher(email).matches()) {
_emailText.setError("enter a valid email address");
valid = false;
} else {
_emailText.setError(null);
}
if (password.isEmpty() || password.length() < 4 || password.length() > 10) {
_passwordText.setError("between 4 and 10 alphanumeric characters");
valid = false;
} else {
_passwordText.setError(null);
}
return valid;
}
}
答案 6 :(得分：0)
private void isValidEmail(String email_id) {
if (email_id == null){
checkTextView.setVisibility(View.VISIBLE);
checkTextView.setText(LocaleController.getString("EnterValidEmail", R.string.EnterValidEmail));
return;
}
if (android.util.Patterns.EMAIL_ADDRESS.matcher(email_id).matches()) {
checkTextView.setVisibility(View.GONE);
} else {
checkTextView.setVisibility(View.VISIBLE);
checkTextView.setText(LocaleController.getString("EnterValidEmail", R.string.EnterValidEmail));
}
使用textView ex.checkTextView并验证电子邮件何时有效，然后textview消失，否则显示消息
答案 7 :(得分：0)
使用此功能验证电子邮件ID：
private boolean validateEmaillId(String emailId){
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(emailId).matches();
}
答案 8 :(得分：0)
我在电子邮件模式中查询了多个电子邮件ID验证和一个电子邮件ID。我使用以下方法解决了这个问题：
public static final String patter_emails="^(\\s*,?\\s*[0-9a-za-z]([-.\\w]*[0-9a-za-z])*@([0-9a-za-z][-\\w]*[0-9a-za-z]\\.)+[a-za-z]{2,9})+\\s*$";
public static final String patter_email="^(\\s*[0-9a-za-z]([-.\\w]*[0-9a-za-z])*@([0-9a-za-z][-\\w]*[0-9a-za-z]\\.)+[a-za-z]{2,9})+\\s*$";
以上这些用于Java和Android中的模式。
使用以下方法检查：
答案 9 :(得分：0)
String emailPattern = "[a-zA-Z0-9._-]+@[a-z]+\\.+[a-z]+";
if(emailId.getText().toString().isEmpty()) {
Toast.makeText(getApplicationContext(),"enter email address",Toast.LENGTH_SHORT).show();
else {
if (emailId.getText().toString().trim().matches(emailPattern)) {
Toast.makeText(getApplicationContext(),"valid email address",Toast.LENGTH_SHORT).show();
} else {
Toast.makeText(getApplicationContext(),"Invalid email address", Toast.LENGTH_SHORT).show();
}
}
答案 10 :(得分：0)
尝试此代码
String emailPattern = "(?:[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*|\"(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21\\x23-\\x5b\\x5d-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])*\")@(?:(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?|\\[(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|[a-z0-9-]*[a-z0-9]:(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21-\\x5a\\x53-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])+)\\])";
if (!tvEmail.matches(Constants.emailPattern)){
tvEmail.setError("Invalid Email ");
}
else{
//your code
}
答案 11 :(得分：0)
String emailPattern = "(?:[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*|\"(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21\\x23-\\x5b\\x5d-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])*\")@(?:(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?|\\[(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|[a-z0-9-]*[a-z0-9]:(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21-\\x5a\\x53-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])+)\\])";
if (!"anan.pp@gmail.co".matches(Constants.emailPattern)){
//tvEmail.setError("Invalid Email ");
} else {
// your code
}答案 12 :(得分：-1)
尝试以下代码：
只需拨打以下方法，
即可if(emailValidator(mail.getText().toString())){
Toast.makeText(getActivity(), "valid email address",
Toast.LENGTH_SHORT).show();
}else{
Toast.makeText(getActivity(), "invalid email address",
Toast.LENGTH_SHORT).show();
}
public static boolean emailValidator(final String mailAddress) {
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,})$";
pattern = Pattern.compile(EMAIL_PATTERN);
matcher = pattern.matcher(mailAddress);
return matcher.matches();
}
答案 13 :(得分：-1)
指定一个String变量来存储此EditText的值：
emailInput = mail.getText().toString().trim();
在EditText中使用setError：
if(!isValidEmail(emailInput)){
mail.setError("Invalid"); /*"Invalid Text" or something like getString(R.string.Invalid)*/
mail.requestFocus();
}
创建检查电子邮件的方法：
private boolean isValidEmail(String emailInput) {
String EMAIL_PATTERN = "^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*@"
+ "[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$";
Pattern pattern = Pattern.compile(EMAIL_PATTERN);
Matcher matcher = pattern.matcher(emailInput);
return matcher.matches();
}