Phone number validation Android
AndroidValidationPhone NumberAndroid Problem Overview
How do I check if a phone number is valid or not? It is up to length 13 (including character + in front).
How do I do that?
I tried this:
String regexStr = "^[0-9]$";
String number=entered_number.getText().toString();
if(entered_number.getText().toString().length()<10 || number.length()>13 || number.matches(regexStr)==false ) {
Toast.makeText(MyDialog.this,"Please enter "+"\n"+" valid phone number",Toast.LENGTH_SHORT).show();
// am_checked=0;
}`
And I also tried this:
public boolean isValidPhoneNumber(String number)
{
for (char c : number.toCharArray())
{
if (!VALID_CHARS.contains(c))
{
return false;
}
}
// All characters were valid
return true;
}
Both are not working.
Input type: + sign to be accepted and from 0-9 numbers and length b/w 10-13 and should not accept other characters
Android Solutions
Solution 1 - Android
Use isGlobalPhoneNumber() method of PhoneNumberUtils to detect whether a number is valid phone number or not.
Example
System.out.println("....g1..."+PhoneNumberUtils.isGlobalPhoneNumber("+912012185234"));
System.out.println("....g2..."+PhoneNumberUtils.isGlobalPhoneNumber("120121852f4"));
The result of first print statement is true while the result of second is false because the second phone number contains f.
Solution 2 - Android
Given the rules you specified:
> upto length 13 and including character + infront.
(and also incorporating the min length of 10 in your code)
You're going to want a regex that looks like this:
^\+[0-9]{10,13}$
With the min and max lengths encoded in the regex, you can drop those conditions from your if() block.
Off topic: I'd suggest that a range of 10 - 13 is too limiting for an international phone number field; you're almost certain to find valid numbers that are both longer and shorter than this. I'd suggest a range of 8 - 20 to be safe.
[EDIT] OP states the above regex doesn't work due to the escape sequence. Not sure why, but an alternative would be:
^[+][0-9]{10,13}$
[EDIT 2]
OP now adds that the + sign should be optional. In this case, the regex needs a question mark after the +, so the example above would now look like this:
^[+]?[0-9]{10,13}$
Solution 3 - Android
To validate phone numbers for a specific region in Android, use libPhoneNumber from Google, and the following code as an example:
public boolean isPhoneNumberValid(String phoneNumber, String countryCode) {
// NOTE: This should probably be a member variable.
PhoneNumberUtil phoneUtil = PhoneNumberUtil.getInstance();
try {
PhoneNumber numberProto = phoneUtil.parse(phoneNumber, countryCode);
return phoneUtil.isValidNumber(numberProto);
} catch (NumberParseException e) {
System.err.println("NumberParseException was thrown: " + e.toString());
}
return false;
}
Solution 4 - Android
You can use android's inbuilt Patterns:
public boolean validCellPhone(String number) {
return android.util.Patterns.PHONE.matcher(number).matches();
}
> This pattern is intended for searching for things that look like they > might be phone numbers in arbitrary text, not for validating whether > something is in fact a phone number. It will miss many things that are > legitimate phone numbers. > > The pattern matches the following: > > - Optionally, a + sign followed immediately by one or more digits. > Spaces, dots, or dashes may follow. > - Optionally, sets of digits in > parentheses, separated by spaces, dots, or dashes. > - A string starting > and ending with a digit, containing digits, spaces, dots, and/or > dashes.
Solution 5 - Android
you can also check validation of phone number as
/**
* Validation of Phone Number
*/
public final static boolean isValidPhoneNumber(CharSequence target) {
if (target == null || target.length() < 6 || target.length() > 13) {
return false;
} else {
return android.util.Patterns.PHONE.matcher(target).matches();
}
}
Solution 6 - Android
You can use PhoneNumberUtils if your phone format is one of the described formats. If none of the utility function match your needs, use regular experssions.
Solution 7 - Android
We can use pattern to validate it.
> android.util.Patterns.PHONE
public class GeneralUtils {
private static boolean isValidPhoneNumber(String phoneNumber) {
return !TextUtils.isEmpty(phoneNumber) && android.util.Patterns.PHONE.matcher(phoneNumber).matches();
}
}
Solution 8 - Android
^\+?\(?[0-9]{1,3}\)? ?-?[0-9]{1,3} ?-?[0-9]{3,5} ?-?[0-9]{4}( ?-?[0-9]{3})?
Check your cases here: https://regex101.com/r/DuYT9f/1
Solution 9 - Android
String validNumber = "^[+]?[0-9]{8,15}$";
if (number.matches(validNumber)) {
Uri call = Uri.parse("tel:" + number);
Intent intent = new Intent(Intent.ACTION_DIAL, call);
if (intent.resolveActivity(getPackageManager()) != null) {
startActivity(intent);
}
return;
} else {
Toast.makeText(EditorActivity.this, "no phone number available", Toast.LENGTH_SHORT).show();
}
Solution 10 - Android
val UserMobile = findViewById<edittext>(R.id.UserMobile)
val msgUserMobile: String = UserMobile.text.toString()
fun String.isMobileValid(): Boolean {
// 11 digit number start with 011 or 010 or 015 or 012
// then [0-9]{8} any numbers from 0 to 9 with length 8 numbers
if(Pattern.matches("(011|012|010|015)[0-9]{8}", msgUserMobile)) {
return true
}
return false
}
if(msgUserMobile.trim().length==11&& msgUserMobile.isMobileValid())
{//pass}
else
{//not valid}
Solution 11 - Android
^\+201[0|1|2|5][0-9]{8}
this regex matches Egyptian mobile numbers
Solution 12 - Android
Here is how you can do it succinctly in Kotlin:
fun String.isPhoneNumber() =
length in 4..10 && all { it.isDigit() }
Solution 13 - Android
I got best solution for international phone number validation and selecting country code below library is justified me Best library for all custom UI and functionality CountryCodePickerProject
Solution 14 - Android
What about this method:
private static boolean validatePhoneNumber(String phoneNumber) {
// validate phone numbers of format "1234567890"
if (phoneNumber.matches("\\d{10}"))
return true;
// validating phone number with -, . or spaces
else if (phoneNumber.matches("\\d{3}[-\\.\\s]\\d{3}[-\\.\\s]\\d{4}"))
return true;
// validating phone number with extension length from 3 to 5
else if (phoneNumber.matches("\\d{3}-\\d{3}-\\d{4}\\s(x|(ext))\\d{3,5}"))
return true;
// validating phone number where area code is in braces ()
else if (phoneNumber.matches("\\(\\d{3}\\)-\\d{3}-\\d{4}"))
return true;
// Validation for India numbers
else if (phoneNumber.matches("\\d{4}[-\\.\\s]\\d{3}[-\\.\\s]\\d{3}"))
return true;
else if (phoneNumber.matches("\\(\\d{5}\\)-\\d{3}-\\d{3}"))
return true;
else if (phoneNumber.matches("\\(\\d{4}\\)-\\d{3}-\\d{3}"))
return true;
// return false if nothing matches the input
else
return false;
}
System.out.println("Validation for 1234567890 : " + validatePhoneNumber("1234567890"));
System.out.println("Validation for 1234 567 890 : " + validatePhoneNumber("1234 567 890"));
System.out.println("Validation for 123 456 7890 : " + validatePhoneNumber("123 456 7890"));
System.out.println("Validation for 123-567-8905 : " + validatePhoneNumber("123-567-8905"));
System.out.println("Validation for 9866767545 : " + validatePhoneNumber("9866767545"));
System.out.println("Validation for 123-456-7890 ext9876 : " + validatePhoneNumber("123-456-7890 ext9876"));
And the outputs:
Validation for 1234567890 : true
Validation for 1234 567 890 : true
Validation for 123 456 7890 : true
Validation for 123-567-8905 : true
Validation for 9866767545 : true
Validation for 123-456-7890 ext9876 : true
For more info please refer to this link.
Solution 15 - Android
Try this function it should work.
fun isValidPhone(phone: String): Boolean =
phone.trimmedLength() in (10..13) && Patterns.PHONE.matcher(phone).matches()
Solution 16 - Android
You shouldn't be using Regular Expressions when validating phone numbers. Check out this JSON API - numverify.com - it's free for a nunver if calls a month and capable of checking any phone number. Plus, each request comes with location, line type and carrier information.