基于判别器的 Typescript 窄参数类型

我有一个旧版 API，如下所示（游乐场链接 https://www.typescriptlang.org/play?#code/C4TwDgpgBAwg9gWwQQwHYBMCMUC8UDeAUFCVAMYLoBcUARAiFAGYCWATgM7DmIoa0AaYqWRsA5phpc2LVGKGkoosQCYaAIzhwANhDSEAvoUKhIsXmnQrcBYSQrU6DKBwhk4GHkkuC7zLVLAMnIKpOqiNKgArgjqEGyGxqbQ8N6eeKl8WFAAPuZpVsZMUahkwCweUBAAHm5RwBAAFA40mZYA2rQOtAC6AkriHFRojAD0AFRQAO5wUdroUNosANbQwHAuQR5i2ozJUMAAFiwcUOOjAJS2iu6oHDoQAHTacGKNyhwXiYQ1dQ2N9EYrE43HcBUEBAmUC0YFOCBOHFkYigYDYcFhZ1GBi+QA)...

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 */})

有什么方法可以输入检查args的参数execute函数不改变函数的参数列表？

Thanks

Use 提取 https://www.typescriptlang.org/docs/handbook/utility-types.html#extracttype-union (操场 https://www.typescriptlang.org/play?#code/C4TwDgpgBAwg9gWwQQwHYBMCMUC8UDeAUFCVAMYLoBcUARAiFAGYCWATgM7DmIoa0AaYqWRsA5phpc2LVGKGkoosQCYaAIzhwANhDSEAvoUKhIsXmnQrcBYSQrU6DKBwhk4GHkkuC7zLVLAMnIKpOqiNKgArgjqEGyGxqbQ8N6eeKl8WFAAPuZpVsZMUahkwCweUBAAHm5RwBAAPDBV1Q0YHPlZANq0DrQAugB8ABQONDACSuIcNADyCCzAjQCibWzIZc0WGFP45JQTUAZDU32UtEMAlLaK7qgcOhAAdNpwYmOUU8ocV4mENTqDRG9EYrE43HcBUEBGmEhotAA1hx0DDlGooEEotADH8AbUyPUICDnK57ugvFkYfsmAE6ABHADuEBh4TYNAALNZcYQgA):

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 })