check telephone number format in javascript

Question

I want to use regular expression in JavaScript which limit number of phone in ten digit and the two first digit must begin with 06 or 07 .

• Example correct format: 0688342134 , 0721654709 ;
• Example bad format: 0588342134 , 0421654709 .

var phoneno = /^\+?([0-9]{2})\)?[-. ]?([0-9]{4})[-. ]?([0-9]{4})$/;
if (inputtxt.value.match(phoneno)) {
  return true;
} else {  
  alert("valid phone number must begin with 06 or 07 and have ten digit");
  return false;
}

It doesn't begin with 06 or 07 .
Any ideas?

Answer 1

 function checkPhone(phone) { return (new RegExp(/^0[6-7][\d]{8}$/)).test(phone); } let phones = [ '0678901234', // true: starting with '06' and 10-chars length '067890123', // false: starting with '06' but 9-chars length '06789012345', // false: starting with '06' but 11-chars length '0789012345', // true: starting with '07' and 10-chars length '0890123456' // false: 10-chars length but starting with '08' ]; phones.forEach(phone => { console.log(checkPhone(phone)); });

Answer 2

you can use

    <input type="tel" pattern="06|07\d{8}">

https://developer.mozilla.org/en-US/docs/Web/HTML/Element/input/tel

Answer 3

this is what you want:

reg = /^0[67]\d{8}$/

• ^: # Assert position at the beginning of the string.
• 0[67]: means 0 followed by 6 or 7.
• d{8}: means 8 numbers
• $: # Assert position at the end of the string.

Answer 4

if(inputtxt.value.match(phoneno) && (inputtxt.value.substring(0,2)=='06' || inputtxt.value.substring(0,2)=='07')) { return true; }

Answer 5

Update your regular expression to:

var phoneno = /^\+?(0[67])\)?[-. ]?([0-9]{4})[-. ]?([0-9]{4})$/;

The part that matches the first two numbers changed from ([0-9]{2}) , which matches 2 of any number, to (0[67]) , which matches a 0 followed by a 6 or 7.

Answer 6

You can use this:

var phoneno = /^\+?(0[67])\)?[-. ]?([0-9]{4})[-. ]?([0-9]{4})$/

Change

[0-9]{2}

To

0[67]

Explanation

0 matches the digit zero.

[67] matches one occurrence of either 6 or 7.