typescript - 如何检查强类型对象是否包含 TypeScript 中的给定键而不提示？

Question

拿这个代码:

const lookup = {
    foo: 1,
    bar: 2
}

const getValueOrDefault = (name: string, defaultValue: number) => {
    if (name in lookup) {
        return lookup[name] // ERROR
    }

    return defaultValue
}

lookup[name]表达式导致此 TS 错误( playground ):

Element implicitly has an 'any' type because expression of type 'string' can't be used to index type '{ foo: number; bar: number; }'.
  No index signature with a parameter of type 'string' was found on type '{ foo: number; bar: number; }'.

...即使我做了 if (name in lookup)先检查一下。我试过 hasOwnProperty ，这也不起作用。

TypeScript 通常很聪明地在条件块中细化类型，但在这里不是。为什么？我怎样才能让它工作(而不仅仅是通过放松 lookup 的类型来破解它)？

Answer 1

您可以包装 name in lookup入住type guard帮助 typescript 理解一旦条件为真 - name有效 lookup的关键:

const getValueOrFalse = (name: string) => {
    if (isObjKey(name, lookup)) {
        // name narrowed to "foo" | "bar"
        return lookup[name]
    }

    return false
}

function isObjKey<T>(key: any, obj: T): key is keyof T {
    return key in obj;
}

Playground