JavaScript - super() keyword unexcepted

Question

I'm getting in trouble with a small problem and I wish that somebody can help me resolving this fast. Here is my code :

class Client {
    /**
   * @param {ClientOptions} [options] Options for the client
   */
  constructor(options = {}) {
    super();
    this.lnk = "https://marketpala.glitch.me";
    this.endpoints = [
        "/commandes.json",
        "/users.json",
        "/blacklist.json",
        "/infov.json",
        "/reserved.json"
    ]
  }
}

Here is the code to export the client :

module.exports = {
    Client: require('./client/Client')
}

Here is the code I use to test my client :

const tst = require('./palamazon')
let t = new tst.Client()

And here is the error I get :

super();
    ^^^^^

SyntaxError: 'super' keyword unexpected here

Hope somebody can help me!

(I'm coding in javascript)

Does this answer your question? [super keyword unexpected here](https://stackoverflow.com/questions/39263358/super-keyword-unexpected-here) — Craicerjack, Mar 03 '20 at 10:28
@KoroBaka If you're following some kind of tutorial and it told you to write super(), it is highly likely you were supposed to extend Client from something. In fact based on the way you're using the class (just overwriting the constructor and adding nothing else) I'd say that it is even likely. — Michael, Mar 03 '20 at 11:05

Nitheesh · Accepted Answer · 2020-03-03T10:36:48.683

4

Super keyword is used in inherited classes to use their properties inside the child class. Your class is not extended from any other classes, so super is not accepted.

By calling the super() method in the constructor method, we call the parent's constructor method and gets access to the parent's properties and methods:

If your class is not being extended from any other classes, the super method must be deleted.

Check some details here.

edited Mar 03 '20 at 10:36

answered Mar 03 '20 at 10:29

Nitheesh

19,238
3
22
49

If the class is not extending any other base class `super` should be deleted. – Nitheesh Mar 03 '20 at 10:33
Yes, there is no need to call super – Ajay Kumar Mar 03 '20 at 10:34
3

It's like trying to call your parents if you're an orphan. – Raekh Void Mar 03 '20 at 10:34
@RaekhVoid the best example that we could give :) – Nitheesh Mar 03 '20 at 10:35
TypeError: tst.Client is not a constructor – koro Mar 03 '20 at 10:36
what is this `let t = new api.Client()`? You have not imported or declared api. – Nitheesh Mar 03 '20 at 10:38
It's tst.client, I just failed in the main post – koro Mar 03 '20 at 10:39
Yup, Client: require('./client/Client') – koro Mar 03 '20 at 10:42
Ok basically I'm just an idiot, I forgot module.exports = Client; – koro Mar 03 '20 at 10:44
Hehe . Happy coding – Nitheesh Mar 03 '20 at 10:45

score 0 · Answer 2 · answered Mar 03 '20 at 10:32

0

As you are not extending any class so the super is not expected, super calls parent class constructor but in this case there is no parent to this class.

answered Mar 03 '20 at 10:32

Ajay Kumar

99
6

score 0 · Answer 3 · edited Mar 03 '20 at 10:43

super() is used to call the original method on the parent class when you are extending one class from another like so,

class MyFirstTestClass {
    constructor() {
        console.log("Hello");
    }
}

class MySecondTestClass extends MyFirstTestClass {
    constructor() {
        super();

        console.log("World");
    }
}

const test = new MySecondTestClass();

This will output Hello and then World. Without calling super(), only World would be outputted because the constructor on the second class would override the constructor on the first.

The class you wrote does not extend from another class, and so super() has no parent class to call to.

You should be able to fix your problem by either inheriting from another class if that was your intention, or by simply deleting the line where you call super().

JavaScript - super() keyword unexcepted

3 Answers3