我有一个联合类型的函数:
function Function1(arg0: string, arg1: any[], name: "hello" | "bye") {
return name;
}
function Function2(arg0: string, arg1: any[], name: "foo" | "bar") {
return name;
}
type Functions = typeof Function1 | typeof Function2;
现在我想创建一个类型，它接受一个元组，其中第一个位置是实际函数，第二个位置是相应的"name"参数.
const baz = [Function1, "foo"];
我试过这个:
type FooBar<T = Functions> = [T, Parameters<T>[2]];
首先，它给出一个错误Parameters<T>[2]"类型‘T’不满足约束‘(...args:any)=>；any’"
"
它几乎可以工作，但我从两个函数中获得了所有参数.
你想要distribute你的[T, Parameters<T>[0]]运算超过T的unions，所以我们可以写一个distributive conditional type来做到这一点.让我们来介绍一下实用程序类型:
type FuncThirdArgTuple<T> =
T extends (a1: any, a2: any, a3: infer A) => any ? [T, A] : never
它不使用Parameters<T>[2]，而是直接使用infer来获取第三个参数类型.
如果我们将它应用于Functions，我们得到:
type FuncThirdArgTupleForFunctions = FuncThirdArgTuple<Functions>
/* type FuncThirdArgTupleForFunctions =
[
(arg0: string, arg1: any[], name: "hello" | "bye") => "hello" | "bye",
"hello" | "bye"
] | [
(arg0: string, arg1: any[], name: "foo" | "bar") => "foo" | "bar",
"foo" | "bar"
]
*/
看起来像是你想要的类型.让我们来测试一下:
let baz: FuncThirdArgTupleForFunctions;
baz = [function1, "hello"]; // okay
baz = [function1, "foo"]; // error
看上go 不错.