I need to validate number that use inputs into input box. Number must be exactly 12 characters long and start with 931.

var reg = /^931[0-9]$/;
var val = $("#sms_number").val();

if (val.length === 12 && reg.test(val)) {                    


}

Above is incorrect.

I need to validate number that use inputs into input box. Number must be exactly 12 characters long and start with 931.

var reg = /^931[0-9]$/;
var val = $("#sms_number").val();

if (val.length === 12 && reg.test(val)) {                    


}

Above is incorrect.

• javascript
• jquery
• regex

• The main problem here is that [0-9] means "any digit between 0 and 9", but repeated just once. Your code would work find if you used [0-9]+ (meaning "any digit between 0 and 9, repeated 1 or more times") – h2ooooooo Commented May 20, 2013 at 11:55

3 Answers 3

var reg = /^931\d{9}$/;
var val = $("#sms_number").val();
if (reg.test(val)) {
    // ....
}

You can do this without regex too.

var val = $("#sms_number").val();

if (val.length === 12 && val.toString().indexOf("931") == 0) {                    

}

Simple but probably not as robust as regex solutions

In a concise way the following should work:

/^931[0-9]{9}$/.test($("#sms_number").val())? console.log(true): console.log(false);

xdazz's answer is correct, but I'd suggest not to use \d if you are not trying to acmodate unicode numbers as well. If you are fine with only English numbers (I think that's the way it meant to be, as it uses 931 prefix which is indeed in English) you should use [0-9] instead of \d. Former is faster.