I want to create a type for an object with string keys and arbitrary values. Thus, my desired type (let's call it Obj) should cause a type error for the following obj object because its key is of type number:
const obj: Obj = {
1: "foo" // I'd expect a TS error here because the key 1 is of type `number` and not `string`
};
However, all of my attempts don't cause any TS error! Here are my three attempts:
- Built-in
Recordtype (which implicitly uses theinoperator):
type Obj = Record<'1', any>;
const obj: Obj = {
1: "foo" // TS error expected! ❌
};
- Mapped type (to recreate
Recordtype):
type Obj = { [key: string]: any };
const obj: Obj = {
1: "foo" // TS error expected! ❌
};
- Custom
Recordtype with key remapping viaas:
type MyRecord<K extends string, T> = {
[P in K as string]: T;
};
type Obj = MyRecord<string, any>;
const obj: Obj = {
1: "foo" // TS error expected! ❌
};
- Even using a literal type
'1'does not enforce the key to be'1', but allows the number1:
type Obj = Record<'1', any>;
const obj: Obj = {
1: "foo" // TS error expected! ❌
};
Is it in general possible, or is it due to some property of a JavaScript object that it's not possible?
CodePudding user response:
The 1 is coerced to a string ("1") when defining an object literal. You cannot define an object with number keys:
const obj = {
prop: 'value',
1: 'another value',
};
for (const key in obj) {
console.log(key, typeof key);
}
CodePudding user response:
By the keyof page of TypeScript doc
JavaScript object keys are always coerced to a string, so obj[0] is always the same as obj["0"]
So in fact {1: "foo"} is just as same as {"1": "foo"}.
