Typescript：将类型定义为“任何类型除外”而不触发错误？

Question

I have a type for any plain JS object:

type Entity = {
  [k: string]: any,
};

I'm incrementally adding TS to my codebase, so I'm allowing implicit any. With this type, I can do:

const user: Entity = ...;
user.name.includes(...);

However, if I want to define it as any plain JS object without object or array values:

type Entity = {
  [k: string]: string | number | boolean | null,
};

I get: Property 'includes' does not exist on type 'string | number | boolean'. Property 'includes' does not exist on type 'string | number | boolean'.

I also tried this, not sure if it's valid TS:

type Entity = {
  [k: string]: Entity[k] extends Object ? never : Entity[k];,
};

But same thing: Property 'includes' does not exist on type 'never'.

Is it possible to get TS to not throw errors in this case? I want it to behave as if the object value's type was any .

Answer 1

Unless someone comes up with a really clever way of working around Object and any and before not object works, I'd recommend trying to build something like

type Entity = {
  name?: string;
  [key: string]: string | boolean | number | undefined;
}

type Entity = {
  name?: string;
  [key: string]: any;
}

Hopefully, you can narrow down your types. In the meantime, I'd try to add more precision and safety where possible. If Entity comes from an external source, then you might want to consider using unknown type.