Typescript基于鉴别器的窄参数类型

Question

我有一个遗产 API 如下所示 (playground link)...

type Command1 = {
    cmd: "my first command",
    arg1: string,
    arg2: boolean
}

type Command2 = {
    cmd: "my second command",
    foo: string,
    bar: number
}

type Command = Command1 | Command2

function execute(cmd: Command["cmd"], args:any /* would like to strongly type this */) {
    console.log(args)
}

execute("my first command", {/* oops missing props */})

有没有什么方法可以在不更改函数参数列表的情况下对 execute 函数的 args 参数进行类型检查？

谢谢

Answer 1

为此使用带有泛型的交集类型怎么样？我假设其余代码与您的示例相同：

function execute<T extends Command["cmd"]>(cmd: T, args: Command & { cmd: T }) {
    console.log(cmd, args)
}

我想不出更好的方法将鉴别器和 Command 的正确类型“粘合”在一起。

Answer 2

使用Extract<Type, Union> (playground):

function execute<Cmd extends Command["cmd"]>(
  cmd: Cmd, 
  args: Omit<Extract<Command, { cmd: Cmd }>, "cmd">
) {
    console.log(cmd, args)
}

// execute<"my first command">(cmd: "my first command",   args: Omit<Command1, "cmd">): void
execute("my first command", { arg1: "ksd", arg2: true })

// execute<"my second command">(cmd: "my second command", args: Omit<Command2, "cmd">): void
execute("my second command", { foo: "qwe", bar: 123 })

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