typescript - Using union type with generic

Question

I have a function that accepts a parameter which is an object consisting of a field (string) and arguments (object literal).

I'd like the arguments type check to be validated dependent on what the field is. I also wish to make a definition that can be extended wherever I needed (hence the generic).

Definition:

 export interface EventDefinition<
    TField extends string,
    TArgs extends any
  > {
    field: TField
    arguments: TArgs
  }

My function:

export const myFn =  (
  params:
    | EventDefinition<
        'actionOne',
        {
          id: string
        }
      >
    | EventDefinition<
        'actionTwo',
        {
          emailAddress: string
        }
      >
) => {
  const { args, field } = params

  switch(field) {
    case 'actionOne':
      console.log(args.id)
      break;
      case 'actionTwo':
      console.log(args.emailAddress)
      break;
  }
}

While the field property validates the args do not and result in errors (for ex with args.id):

Property 'id' does not exist on type '{ id: string }'.
Property 'id' does not exist on type '{ emailAddress: string; }'.

How can I correctly define this?

标签： typescript

Answer 1

Typescript will not guard one variable based on another, this is just not supported.

If you use the argument itself control flow analysis can determine the types appropriately

  switch(params.field) {
    case 'actionOne':
      console.log(params.arguments.id)
      break;
      case 'actionTwo':
      console.log(params.arguments.emailAddress)
      break;
  }

Playground Link