Return everything after phrase, to end of string, excluding phrase

Question

I have the following URL

https://website.com?id=XXXVVVCCCHHH

I only want XXXVVVCCCHHH, I've tried the following:

var phrase = 'https://website.com?id=XXXVVVCCCHHH'; 
var myRegexp = /id=(.*)/;
phrase = myRegexp.exec(phrase);

But this is returning: id=XXXVVVCCCHHH;

How can Ii edit this to only return XXXVVVCCCHHH?

Your solution is already working well. Just access the right group value. — Wiktor Stribiżew, Oct 24 '17 at 14:41
`id=(.*?)(?=$|[])` in the case another get variable or fragment may be present — ctwheels, Oct 24 '17 at 14:45

score 2 · Accepted Answer · answered Oct 24 '17 at 14:38

2

Just use split and take the second element:

var url = "https://website.com?id=XXXVVVCCCHHH";
var part = url.split('=')[1];
console.log(part);

answered Oct 24 '17 at 14:38

Faly

1 Answers1