EditText 上的 Android 电子邮件验证
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时间：2020-08-20 08:58:41 来源:igfitidea点击:
Android Email Validation on EditText
提问by BekaKK
I have one edittext and I would to to write email validation in my Editttext this is a xml code
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>
and this is a java code
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);
}
I can't validation. The toast message is always "Invalid email address". What am I doing wrong?
我无法验证。Toast 消息始终是“无效的电子邮件地址”。我究竟做错了什么？
回答by Daniel Zolnai
回答by sachin pangare
I am posting very simple and easy answer of email validation without using any string pattern.
我在不使用任何字符串模式的情况下发布了非常简单易行的电子邮件验证答案。
1.Set on click listener on button....
1.在按钮上设置单击侦听器....
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.call the isValidEmail() i.e..
2.调用 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();
}
}
I hope it will helpful for you...
我希望它对你有帮助......
回答by Raseem Ayatt
Android Email Validation Simplest way
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();
}
回答by Shamir Kp
Try following code:
尝试以下代码：
public final static boolean isValidEmail(CharSequence target) {
return !TextUtils.isEmpty(target) && android.util.Patterns.EMAIL_ADDRESS.matcher(target).matches();
}
This works fine.
这工作正常。
回答by ivap arht
android.util.Patterns.EMAIL_ADDRESS.matcher(target).matches()
回答by Tabish khan
here is a whole code for login validations......
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;
}
}
}
回答by rajeev omar
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));
}
Take a textView ex.checkTextView and validate when the email is valid then the textview is gone otherwise it show the message
使用 textView ex.checkTextView 并验证电子邮件何时有效然后 textview 消失，否则它会显示消息
回答by Vishal Vaishnav
Use this function for validating email id:
使用此函数验证电子邮件 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();
}
回答by ankit
I had a query in email pattern for more than one Email ID validation and for only one Email ID.I solved it by using :
我在电子邮件模式中查询了多个电子邮件 ID 验证和仅一个电子邮件 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*$";
public static final String patter_email="^(\s*[0-9a-za-z]([-.\w]*[0-9a-za-z])*@([0-9a-za-z][-\w]*[0-9a-za-z]\.)+[a-za-z]{2,9})+\s*$";
These above is used for pattern in Java and Android both.
以上这些都用于 Java 和 Android 中的模式。
check this by using:
使用以下方法检查：
回答by rhaldar
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();
}
}
