RegExp in JavaScript is returning null

Question

I want check phone numbers. Right format is 38xxxxxxxxxx.

My check:

var phoneNumber=document.getElementById('phone').value;
var re=new RegExp("^[38]\d{10}$");
var res=phoneNumber.match(re);

I always get null. What's wrong?

Answer 1

When using the RegExp constructor function with quotes, normal string escape rules apply. Thus, you need to escape the special character \\d as \\\\d . Also you need to change [38] to simply 38 , as [38] matches 3 or 8 .

 var str = '381234567890'; var re = new RegExp("^38\\\\d{10}$"); // or new RegExp(/^38\\d{10}$/); without quotes // or re = /^38\\d{10}$/; var res = str.match(re); document.body.innerHTML = "Match result: " + res; 

Answer 2

var res=phoneNumber.match(/38[0-9]{8,10}/m);

phone number length allowed :10-12

Answer 3

您的正则表达式是错误的，如果要匹配38xxxxxxxxxx ，则应删除[]括号，因为这表示3或8 ，然后它会尝试匹配10位数字，因此只需删除[]

var re=new RegExp("^38\d{10}$");

Answer 4

使用构造函数时，必须使用常规的字符串转义规则。

new RegExp("^38\\d{10}$");