typescript :另一个参数的键？

Question

我正在创建一个适用于任何对象的函数，如下所示:

function deleteKey (obj, key) {
  // This is just for an example, but you will get what kind of typing needed.
  delete obj[key];
}

如何正确输入 Typscript？有没有好方法使用 keyof 这样的参数？

function deleteKey (obj: object, key: keyof obj) {
  // This is just for an example, but you will get what kind of typing needed.
  delete obj[key];
}

Answer 1

像这样的事情应该可以解决问题:

function deleteKey<T, K extends keyof T>(obj: T, key: K): Omit<T, K> {
    delete obj[key];
    return obj;
}

interface Foo {
    a: string;
    b: number;
    c: boolean;
}

const foo: Foo = { a: 'test', b: 12, c: true };
const foo_minus_a = deleteKey(foo, 'a');
const foo_minus_b = deleteKey(foo, 'b');
const foo_minus_c = deleteKey(foo, 'c');