How to set type in Factory pattern in which exists multiple interfaces

Question

There are like these types

interface A {
  id: number
  name: string
  patternA: patternA[]
}

interface B {
  id: number
  name: string
  patternB: patternB[]
}

interface C {
  id: number
  name: string
  patternC: patternC[]
  city: string
}

and there is factory method

const factory = (type, data: /** 🤔 what should I write?? **/) => {
  switch (type) {
    case 'first':
      return createHogeA(data)
    case 'second':
      return createHogeB(data)
    case 'third':
      return createHogeC(data)
    default:
      return
  }
}

const createHogeA = (data: A) => somefunctionA(data)
const createHogeB = (data: B) => somefunctionB(data)
const createHogeC = (data: C) => somefunctionC(data)

What I've tried so far

Set type like this, but then appears error because it's possible that each datas type could pass to each functions

const factory = (type, data: A | B | C) => {
  switch (type) {
    case 'first':
      return createHogeA(data)
    case 'second':
      return createHogeB(data)
    case 'third':
      return createHogeC(data)
    default:
      return
  }
}

Create base interface

interface base {
  id: number
  name: string
}

But then honestly I didn't have idea

how do you intend to call your factory? where does type come from? — Patrick Michaelsen
– Patrick Michaelsen, Commented Mar 30, 2023 at 7:05

Patrick Michaelsen · Accepted Answer · 2023-03-30 07:07:41Z

0

Perhaps via type union discrimination:

interface A {
  type: 'A'
  id: number
  name: string
  patternA: patternA[]
}

interface B {
  type: 'B'
  id: number
  name: string
  patternB: patternB[]
}

interface C {
  type: 'C'
  id: number
  name: string
  patternC: patternC[]
  city: string
}

type ABC = A | B | C;

const factory = (data: ABC) => {
  switch (data.type) {
    case 'A':
      return createHogeA(data)
    case 'B':
      return createHogeB(data)
    case 'C':
      return createHogeC(data)
    default:
      return
  }
}

answered Mar 30, 2023 at 7:07

Patrick Michaelsen

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Comments

Milly Stack · Accepted Answer · 2023-03-30 06:52:07Z

-1

The easiest way to handle this is to use the as keyword to tell the compiler which of the interfaces the data object satisfies when calling either of the functions.

const factory = (type: string, data: A | B | C) => {
  switch (type) {
    case 'first':
      return createHogeA(data as A)
    case 'second':
      return createHogeB(data as B)
    case 'third':
      return createHogeC(data as C)
    default:
      return
  }
}

You could also use guards to determine which interface data satisfies

const isA = (data: any): data is A => {
  return (data as A).patternA !== undefined;
}

const isB = (data: any): data is B => {
  return (data as B).patternB !== undefined;
}

const isC = (data: any): data is C => {
  return (data as C).patternC !== undefined && (data as C).city !== undefined;
}

const factory = (_type: string, data: A | B | C) => {
  if(isA(data)) {
    createHogeA(data)
  }
  // Rest of code here
}

answered Mar 30, 2023 at 6:52

Milly Stack

1395 bronze badges

1 Comment

kazon Over a year ago

I've tried it and it resolved, thank you :)

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