TS2339 - 明显有效的 TS 文件中的“类型上不存在属性"错误 [英] TS2339 - 'Property doesn't exist on type' error in an apparently valid TS file

问题描述

我有以下类型声明:

type Root = { r: string };
type A = { a: string };
type B = { b: string };
type Main = Root & (A | B);

由于 Main 等价于 {r: string, a: string} |{r: string, b: string}，这是有效的:

Since Main is equivalent to {r: string, a: string} | {r: string, b: string}, this works:

const main: Main = {
    r: 'r',
    a: 'a'
}

这里没有惊喜.但是，这会引发以下错误:

No surprises here. However, this throws the following error:

const func : (main: Main) => void = main => {
    main.a
    /*   ^
    * [TS2339]
    * Property 'a' doesn't exist on type 'Main'.
    *   Property 'a' doesn't exist on type 'Root & B' */
}

我知道 .a 不存在于 Root &B，但它存在于 Root &A，所以它必须存在于 (Root & A) |(Root & B) 等价于 Main 吧?

I do understand that .a doesn't exist on Root & B, but it exists on Root & A, so it must exists on (Root & A) | (Root & B) which is equivalent to Main, right?

这是一个错误，还是我遗漏了什么?

Is it a bug, or am I missing something?

推荐答案

Main 等同于 (Root & A) | 你说得对(Root & B) 你对联合类型 (|) 的解释是错误的.联合类型意味着您可以将联合中的任何一种类型分配给 main (所以要么 (Root & A) 要么 (Root & B))，但您只能访问联合成员的公共属性.在这种情况下，公共成员只是 r.由于 main 可以是 (Root & A) 或 (Root & B)，a 可能不是存在于 main 上，所以 typescript 阻止了这样的访问.

You are right about Main being equivalent to (Root & A) | (Root & B) you are wrong about the interpretation of union types (|). A union type means you can assign either type in the union to main ( so either (Root & A) or (Root & B)), but you can only access common properties of the union members. In this case the common member is only r. Since main ca be either (Root & A) or (Root & B), a might not exist on main, so typescript prevents such an access.

要使用仅存在于联合中的一种类型的成员，您需要一个类型保护.在这种情况下，in 类型保护将最有效:

To use the members that only exist on one type in the union you need a type guard. In this case an in type-guard will work best:

const func : (main: Main) => void = main => {
    if ('a' in main) {
        main.a
    } else {
        main.b
    }
}

您可以在此处

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