TypeScript: pass generic type as parameter in generic class

Question

TypeScript: I have a method in the DataProvider class with a method getTableData:

public static getTableData<T extends DataObject>(type: { new(): T}): Array<T> { ... }

this works perfectly when I code:

let speakers = DataProvider.getTableData(Speaker);  // where Speaker is a class

now I want to call this from a generic Class:

export class ViewModelBase<T extends DataObject> {   
  public getData(): Array<T> {
    return <T[]> DataProvider.getTableData(T);
  }
}

Now I get a Cannot find name 'T' error for the T parameter I pass to getTableData. How should getTableData be called?

update: With the help of @Paleo I came up this:

export class ViewModelBase<T extends DataObject> {   

  constructor(private dataObjectClass: { new(): T}){}

  public getTableData(): Array<T> {
    return <T[]> DataProvider.getTableData<T>(this.dataObjectClass);
  }
}

the thing is that although I have already told in: class SpeakerViewModel extends ViewModelBase<Speaker> { ... } that I want it to be a ViewModel for Speaker I still have the instantiate the SpeakerViewModel like:

let vm = new SpeakerViewModel(Speaker);

although I have already told it is all about Speaker. I guess I still don't fully understand this.

In DataProvider.getTableData(T), the parameter T is defined nowhere. — Paleo
– Paleo, Commented May 27, 2016 at 13:14
"that I want it to be a ViewModel for Speaker I still have the instantiate the SpeakerViewModel like:let vm = new SpeakerViewModel(Speaker);" sorry, I am confused. Can you rephrase that? What is the code you want to omit? — Aluan Haddad
– Aluan Haddad, Commented Apr 24, 2017 at 23:53
I'm not sure what changed, but I was only able to get this to work with the parameter signature like { new(...args: any[]): T }. I have a suspicion that it's because my class constructor had parameters. — Erik Philips
– Erik Philips, Commented Dec 1, 2019 at 0:20

Paleo · Accepted Answer · 2016-05-27 16:41:54Z

6

Generics are just metadata. They cannot be used as parameters when calling a function. Maybe you need something like this:

export class ViewModelBase<T extends DataObject> {
  constructor(private Cl: {new(): T}) {
  }
  public getData(): Array<T> {
    return DataProvider.getTableData<T>(this.Cl);
  }
}

edited May 27, 2016 at 16:41

answered May 27, 2016 at 13:19

Paleo

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2 Comments

Bronco Oostermeyer Over a year ago

almost but your comment made me figure what I actually needed: constructor(private Cl: { new(): T}).... Thanks!

Paleo Over a year ago

@BroncoOostermeyer OK. I edited to fix the code for future readers.

flags · Accepted Answer · 2018-02-18 14:19:41Z

3

maybe this would help:

export abstract class BaseEntity {
  public static from<T extends BaseEntity>(c: new() => T, data: any): T {
    return Object.assign(new c(), data)
  }
  public static first<T extends BaseEntity>(c: new() => T, data) {
    if (data.rows.length > 0) {
      let item = data.rows.item(0);
      return BaseEntity.from(c, item);
    }
    return null;
  }

}

This class can be extended by others so you could call methods on the base class or on its subclasses.

For instance:

return Product.first(Product, data);

Or:

return BaseEntity.first(Product, data);

See how from() method is called from inside first()

answered Feb 18, 2018 at 14:19

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Comments

glowkeeper · Accepted Answer · 2018-12-03 12:28:58Z

-2

How about defining a base type and extending it? Then your function could expect the base type, and you could call it with the extended type. e.g:

export interface BaseData {
  key: object
}

Then:

import { BaseData } from 'baseDataFile'

export interface DerivedData extends BaseData {
  key: someObjectType
}

Now:

import { BaseData } from 'baseDataFile'

export const someFunc = (props: BaseData) => {
    // do some stuff
    return something 
}

Finally:

import { DerivedData } from 'derivedDataFile'

const myData: DerivedData = something as DerivedData
const myNewData = someFunc(myData)

answered Dec 3, 2018 at 12:28

glowkeeper

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