TypeScript const 断言：如何使用 Array.prototype.includes？

Question

我正在尝试使用一个元素数组作为联合类型，这在 TS 3.4 中使用 const 断言变得很容易，所以我可以这样做：

const CAPITAL_LETTERS = ['A', 'B', 'C', ..., 'Z'] as const;
type CapitalLetter = typeof CAPITAL_LETTERS[string];

现在我想测试一个字符串是否是大写字母，但是以下失败 "not assignable to parameter of type":

let str: string;
...
CAPITAL_LETTERS.includes(str);

有没有比将 CAPITAL_LETTERS 转换为 unknown 然后再转换为 Array<string> 更好的方法来解决这个问题？

Answer 1

The standard library signature for Array<T>.includes(u) 假定要检查的值与数组元素 T 的类型相同或更窄。但是在您的情况下，您正在做相反的事情，检查 wider 类型的值。事实上，只有在 T 和 U 之间没有重叠（即 T & U 是never)。

现在，如果您不打算经常使用 includes() 进行这种 "opposite"，并且您希望零运行时效果，则应该扩大 CAPITAL_LETTERS通过类型断言到 ReadonlyArray<string>：

(CAPITAL_LETTERS as ReadonlyArray<string>).includes(str); // okay

另一方面，如果您认为 includes() 的这种用法应该被接受而没有类型断言，并且您希望它发生在您的所有代码中，您可以 merge in a custom declaration:

// global augmentation needed if your code is in a module
// if your code is not in a module, get rid of "declare global":
declare global { 
  interface ReadonlyArray<T> {
    includes<U>(x: U & ((T & U) extends never ? never : unknown)): boolean;
  }
}

这将使一个数组（好吧，一个只读数组，但这就是你在这个例子中所拥有的）将允许 .includes() 的任何参数，只要数组元素之间有一些重叠类型和参数类型。由于 string & CapitalLetter 不是 never，它将允许调用。不过，它仍然会禁止 CAPITAL_LETTERS.includes(123)。

好的，希望对您有所帮助；祝你好运！

Answer 2

另一种解决方法是使用类型保护

https://www.typescriptlang.org/docs/handbook/advanced-types.html#user-defined-type-guards

const myConstArray = ["foo", "bar", "baz"] as const

function myFunc(x: string) {
    //Argument of type 'string' is not assignable to parameter of type '"foo" | "bar" | "baz"'.
    if (myConstArray.includes(x)) {
        //Hey, a string could totally be one of those values! What gives, TS?
    }
}

//get the string union type
type TMyConstArrayValue = typeof myConstArray[number]

//Make a type guard
//Here the "x is TMyConstArrayValue" tells TS that if this fn returns true then x is of that type
function isInMyConstArray(x: string): x is TMyConstArrayValue {
    return myConstArray.includes(x as TMyConstArrayValue)

    //Note the cast here, we're doing something TS things is unsafe but being explicit about it
    //I like to this of type guards as saying to TS:
    //"I promise that if this fn returns true then the variable is of the following type"
}

function myFunc2(x: string) {
    if (isInMyConstArray(x)) {
        //x is now "foo" | "bar" | "baz" as originally intended!
    }
}

虽然您必须引入另一个“不必要的”功能，但这最终看起来很干净并且工作完美。在你的情况下你会添加

const CAPITAL_LETTERS = ['A', 'B', 'C', ..., 'Z'] as const;
type CapitalLetter = typeof CAPITAL_LETTERS[string];
function isCapitalLetter(x: string): x is CapitalLetter {
    return CAPITAL_LETTERS.includes(x as CapitalLetter)
}

let str: string;
isCapitalLetter(str) //Now you have your comparison

//Not any more verbose than writing .includes inline
if(isCapitalLetter(str)){
  //now str is of type CapitalLetter
}

Answer 3

这是一个适用于使用 TypeScript 4.1 Template Literal Types 的字符串和字符串文字的解决方案，它不会破坏任何其他内容，并且在条件中使用时为了方便还缩小了类型：

declare global {
    interface ReadonlyArray<T> {
        includes<S, R extends `${Extract<S, string>}`>(
            this: ReadonlyArray<R>,
            searchElement: S,
            fromIndex?: number
        ): searchElement is R & S;
    }
}

最初由 noppa 发表于 a TypeScript github issue related to this。

Answer 4

使用 lodash

const CAPITAL_LETTERS = ['A', 'B', 'C', 'Z'] as const;
_.includes(CAPITAL_LETTERS, 'A');

TypeScript const 断言：如何使用 Array.prototype.includes？

TypeScript const assertions: how to use Array.prototype.includes?

typescript3.0