CODEWITHSUNDEEP
typescriptoopinheritance
Greg Wozniak
Greg Wozniak

Reputation: 7172

TS - Cannot call a public method within super() from the children

I've got two classes. One is parent Model class. Another is its children, so that:

import { IModel } from '../shared'

export default class DataModel implements IModel {

  private data: {[key: string]: any} = {}

  public constructor(data: {[key: string]: any}) {
    this.data = data
  }

  // ... other methods

  public forRequest = (params?: {[key: string]: any}): object => {
    return {
      ...this.data,
      ...params
    }
  }
}

and 

import { IModel } from '../shared'
import DataModel from './_'

export default class LoginModel extends DataModel implements IModel {

  constructor() {
    super({
      email: '[email protected]',
      password: 'test1234'
    })
  }

  public forRequest = (): object => {
    return super.forRequest({
      sso: true
    })
  }
}

and:

export interface IModel {
  forRequest(params?: {[key: string]: any}): object
  ...
}

Following the https://www.typescriptlang.org/docs/handbook/classes.html and example with animals, I would like to call my forRequest() method, passing parameters to the parent class.

However, when I am calling super.forRequest(...) I am getting error: Only public and protected methods of the base class are accessible via the 'super' keyword.ts(2340) .

(I have no problem to overload forRequest() method in parent class if this could be a solution, if necessary but not sure if this is such a good idea). (in IModel I tried in both ways, with params and without)

I am wondering what is different here than on the example from TS site, and why the code does not work.

Any help appreciated. Thank you.

Upvotes: 2

Views: 324

Answers (1)

tobenna
tobenna

Reputation: 111

The difference here is with the way you declare your forRequest function.

You're assigning a function as the value of the forRequest property as opposed to assigning it to the prototype.

You only have access to super in the child methods.

You should have access if you did this instead.

class DataModel {

  private data: {[key: string]: any} = {}

  public constructor(data: {[key: string]: any}) {
    this.data = data
  }

  // ... other methods

  public forRequest(params?: {[key: string]: any}): object {
    return {
      ...this.data,
      ...params
    }
  }
}

class LoginModel extends DataModel {

  constructor() {
    super({
      email: '[email protected]',
      password: 'test1234'
    })
  }

  public forRequest(): object {
    return super.forRequest({
      sso: true
    })
  }
}

See what it compiles to here

Upvotes: 2

Related Questions