Typescript's Integrated Compatibility of Types

One important concept I need to convey is that if one of these fields exists, then the other must also exist. When these two fields are peers within the same scope, it can be challenging to clearly communicate this connection.

Consider the example of defining a Pipeline:

type Status = 'running' | 'finished' | 'error' | 'inherited'
type Pipeline = Array<{
    name: string
    status: Status
    parent?: string // problematic. only needed for 'inherited'.
}>

In this scenario, how could you effectively demonstrate that the parent field should always be specified when status='inherited'?

typescripttypes

Answer №1

To suit your specific scenario, conditional types can also be considered. See the example below:

type Status = 'running' | 'finished' | 'error' | 'inherited'

type Pipeline<S extends Status> = Array<{
    name: string
    status: S
    parent: S extends 'inherited' ? string : never
}>

declare const a: Pipeline<'inherited'>
declare const b: Pipeline<'finished'>
a[0].parent // string
b[0].parent // never

Answer №2

Instead of considering the "mutually including" of these options, it is more beneficial to view it as "isolating them from" the rest.

type Status = 'running' | 'finished' | 'error' | 'inherited'
type Pipeline = Array<{
    name: string
    status: Exclude<Status, 'inherited'>
} | {
    name: string
    status: 'inherited'
    parent: string
}>

Naturally, for more complex schemas, this approach requires repeating many fields.

To address this, you can consolidate common fields and incorporate them using the & operator.

type Common = {
    name: string
    lots: string
    of: string
    metadata: string
}
type Pipeline = Array<Common & ({
    status: Exclude<Status, 'inherited'>
} | {
    status: 'inherited'
    parent: string
})>

While this aids in type checking, the abstraction may obscure the actual type during error analysis.

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