Valid URL Regular Expression

valid-url-regex.js

var isValidUrl = /^(?:(?:https?|ftp):\/\/)(?:\S+(?::\S*)?@)?(?:(?!10(?:\.\d{1,3}){3})(?!127(?:\.​\d{1,3}){3})(?!169\.254(?:\.\d{1,3}){2})(?!192\.168(?:\.\d{1,3}){2})(?!172\.(?:1[​6-9]|2\d|3[0-1])(?:\.\d{1,3}){2})(?:[1-9]\d?|1\d\d|2[01]\d|22[0-3])(?:\.(?:1?\d{1​,2}|2[0-4]\d|25[0-5])){2}(?:\.(?:[1-9]\d?|1\d\d|2[0-4]\d|25[0-4]))|(?:(?:[a-z\u00​a1-\uffff0-9]+-?)*[a-z\u00a1-\uffff0-9]+)(?:\.(?:[a-z\u00a1-\uffff0-9]+-?)*[a-z\u​00a1-\uffff0-9]+)*(?:\.(?:[a-z\u00a1-\uffff]{2,})))(?::\d{2,5})?(?:\/[^\s]*)?$/i;

var url = "...";
 
console.log(isValidUrl.test(url)) //=> true/false

var url = "https://wwwww.google.com";

console.log(isValidUrl.test(url)) //=> true/false
VM1063:5 true

It is wrong.Do you think it?

@strongant
https://wwwww.google.com is a valid URL. www has no special meaning. It's just a subdomain of google.com, same as images.google.com or shopping.google.com. What you need to do is extract the host from the URL and then do some sort of validation there. Maybe try a DNS lookup for that host (although that's probably a terrible idea.)

Alternatively, don't validate to that level. Be conservative in what you send, be liberal in what you accept.

should it be working for http://localhost:4000 or http://127.0.0.1:9000