Typescript Interface - Possible to make “one or the other” properties required?

Question

Possibly an odd question, but i\'m curious if it\'s possible to make an interface where one property or the other is required.

So, for example...

int         


        

Answer 1

Thanks @ryan-cavanaugh that put me in the right direction.

I have a similar case, but then with array types. Struggled a bit with the syntax, so I put it here for later reference:

interface BaseRule {
  optionalProp?: number
}

interface RuleA extends BaseRule {
  requiredPropA: string
}

interface RuleB extends BaseRule {
  requiredPropB: string
}

type SpecialRules = Array

// or

type SpecialRules = (RuleA | RuleB)[]

// or (in the strict linted project I'm in):

type SpecialRule = RuleA | RuleB
type SpecialRules = SpecialRule[]

Update:

Note that later on, you might still get warnings as you use the declared variable in your code. You can then use the (variable as type) syntax. Example:

const myRules: SpecialRules = [
  {
    optionalProp: 123,
    requiredPropA: 'This object is of type RuleA'
  },
  {
    requiredPropB: 'This object is of type RuleB'
  }
]

myRules.map((rule) => {
  if ((rule as RuleA).requiredPropA) {
    // do stuff
  } else {
    // do other stuff
  }
})