如何在 typescript 中使用联合类型

Question

I'm using typescript on a project and in some parts of it I have to use union types. But I'm getting some weird error messages that I don't know how to deal with. Consider the type below:

type body = {
  [_: string]:
    | 'boolean'
    | 'number'
    | 'string'
    | {
        type: 'boolean' | 'number' | 'string'
        optional?: boolean
      }
    | {
        type: 'array'
        items: 'boolean' | 'string' | 'number'
        optional?: boolean
        [_: string]: any
      }
}

I'm using [_: string]: because I should be able to use any arbitrary key. The value can either be a string indicating a type or can be a object that provides more details. Now consider the following function:

function show(data: body) {
  console.log(data)
}

When I call the above function with the object below:

const data = {
  username: { type: 'string', optional: false },
  address: { type: 'string' },
  city: 'string'
}

Typescript gives the error below:

Argument of type '{ username: { type: string; optional: boolean; }; address: { type: string; }; city: string; }' is not assignable to parameter of type 'body'.

Property 'username' is incompatible with index signature.

Type '{ type: string; optional: boolean; }' is not assignable to type '"string" | "number" | "boolean" | { type: "string" | "number" | "boolean"; optional?: boolean | undefined; } | { [_: string]: any; type: "array"; items: "string" | "number" | "boolean"; optional?: boolean | undefined; }'.

Property 'items' is missing in type '{ type: string; optional: boolean; }' but required in type '{ [_: string]: any; type: "array"; items: "string" | "number" | "boolean"; optional?: boolean | undefined; }'.

How can I solve this? thanks

Answer 1

You can explictly declare what type your property data is

const data: body = {
  username: { type: 'string', optional: false },
  address: { type: 'string' },
  city: 'string'
}

or cast it

const data = {
  username: { type: 'string', optional: false },
  address: { type: 'string' },
  city: 'string'
} as body;

Answer 2

The issue here is with an overly broad inference on the type of the data variable. Take a look at the reproduction here .

If you hover your data variable, you'll note that it was given the inferred type:

{
    username: {
        type: string;
        optional: boolean;
    };
    address: {
        type: string;
    };
    city: string;
}

Breaking down your error message, we see the first line:

Property 'username' is incompatible with index signature.

Indicates that username is the issue here. As you can see from the inferred type, that property has the type {type: string, optional: boolean} .

You might think that this would match to the

{ type: 'boolean' | 'number' | 'string', optional?: boolean }

part of your union, however the critical difference is that this part of your union requires the type property to have one of the literal string types 'boolean' | 'number' | 'string' 'boolean' | 'number' | 'string' 'boolean' | 'number' | 'string' , and does not permit any string (represented by the type string (without quotes)).

This is why it then tries to match to the final piece of your union, which would allow the type: string key/value pair due to the index signature. But this last part of the union also requires an Items property which is what is leading to the specific error you see.

---

So how do you fix it? Well you need to give typescript a hint that the username.type property of your data object is not just any string. Here are some various ways:

• Use an explicit declaration or cast, using the body type, as in Andreas' excellent answer.
• Cast just that property's value: type: 'string' as 'string' . You'd have to do this for all three properties so this is not very ideal in this particular case.
• Declare the entire object as a const: const data = {... } as const
• Or just use the object inline instead of declaring it first:

show({
  username: { type: 'string', optional: false },
  address: { type: 'string' },
  city: 'string'
});