How to set type when computed property is used?-CodePudding

I have two interfaces.

interface Person {
    name: string;
    weight: number;
}

interface Dog {
    name: string;
    mass: number
}

const attribute = isDog ? 'mass' : 'weight';
const player: Person | Dog = "Object with specific type"

if (player[attribute] === 0) return;

I have this error:
TS7053: Element implicitly has an 'any' type because expression of type '"mass" | "weight"' can't be used to index type 'Person | Dog'.

The API can return both Dog or Person and it depends on other parameter. I would like to know what is the best way to tell TS that if player is Person use 'weight' and if the player is Dog use 'mass'. I dont want to use "any" type.

CodePudding user response：

TypeScript is actually helpful here: if, for example, attribute is "mass", but player is a Person, then it will be a legitimate error since mass property doesn't exist on Person instances.

You avoid this error, you need to change the approach. You could add a runtime check of attribute actually existing in player, but I'd argue that you should use the isDog variable (it's just more logical):

const isDog = "mass" in player;
const mass = isDog ? player.mass : player.weight;

if (mass === 0) return;

Try it.