Question

我正在创建一个可与任何对象一起使用的函数：

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？有没有使用keyof这样的参数的好方法？

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

类似这样的事情应该可以解决：

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');

打字稿：另一个参数的键？

1 个答案: