TypeScript：从 object 推断的类型

Question

Is it possible to automatically generate/infer a flexible type from an object literal?

I have the following object:

export const config = {
  name: 'foo',
  bar: true,
} as const;

I want to automatically generate a flexible type based on this. Something which looks like:

type Config = {
  name: string;
  bar: boolean;
}

I have tried type Config = typeof config , but this gives me a type where name must be 'foo' , etc. All properties are fixed to a single value.

I have also tried the following, but got the same result

export type Config = {
  [K in keyof typeof config]: typeof config[K];
};

I could of course write a separate type def, but my actual use case is more complex than the above example.

Answer 1

Ok, turns out my second example works as long as as const is removed from the object literal used as the template. The readonly attribute can be added to the interred type:

export const config = {
  name: 'foo',
  bar: true,
}; // No `as const`

export type Readonly<Config = {
  [K in keyof typeof config]: typeof config[K];
}>; // Make this type read-only