How to validate phone numbers using regex

Better option... just strip all non-digit characters on input (except 'x' and leading '+' signs), taking care because of the British tendency to write numbers in the non-standard form +44 (0) ... when asked to use the international prefix (in that specific case, you should discard the (0) entirely).

Then, you end up with values like:

 12345678901
 12345678901x1234
 345678901x1234
 12344678901
 12345678901
 12345678901
 12345678901
 +4112345678
 +441234567890

Then when you display, reformat to your hearts content. e.g.

  1 (234) 567-8901
  1 (234) 567-8901 x1234

Have you had a look over at RegExLib?

Entering US phone number brought back quite a list of possibilities.

This is a simple Regular Expression pattern for Philippine Mobile Phone Numbers:

((\+[0-9]{2})|0)[.\- ]?9[0-9]{2}[.\- ]?[0-9]{3}[.\- ]?[0-9]{4}

or

((\+63)|0)[.\- ]?9[0-9]{2}[.\- ]?[0-9]{3}[.\- ]?[0-9]{4}

will match these:

+63.917.123.4567  
+63-917-123-4567  
+63 917 123 4567  
+639171234567  
09171234567  

The first one will match ANY two digit country code, while the second one will match the Philippine country code exclusively.

I work for a market research company and we have to filter these types of input alllll the time. You're complicating it too much. Just strip the non-alphanumeric chars, and see if there's an extension.

For further analysis you can subscribe to one of many providers that will give you access to a database of valid numbers as well as tell you if they're landlines or mobiles, disconnected, etc. It costs money.

Here's one that works well in JavaScript. It's in a string because that's what the Dojo widget was expecting.

It matches a 10 digit North America NANP number with optional extension. Spaces, dashes and periods are accepted delimiters.

"^(\\(?\\d\\d\\d\\)?)( |-|\\.)?\\d\\d\\d( |-|\\.)?\\d{4,4}(( |-|\\.)?[ext\\.]+ ?\\d+)?$"

    pattern="^[\d|\+|\(]+[\)|\d|\s|-]*[\d]$" 
    validateat="onsubmit"

Must end with a digit, can begin with ( or + or a digit, and may contain + - ( or )