从有效属性名称列表中键入具有一个且只有一个属性的对象

Question

我们如何在不创建包含每个可能对象的联合类型的情况下对此进行建模？

type ValidRootPropertyNames =
  'Property1' |
  'Property2';

type ObjectWithRootPropertyNameFromList =
  { Property1: string; } |
  { Property2: string; };

const obj1: ObjectWithRootPropertyNameFromList = { // okay
  Property1: 'foo',
};

const obj2: ObjectWithRootPropertyNameFromList = { // okay
  Property2: 'foo',
};

const obj3: ObjectWithRootPropertyNameFromList = { // error
  Property3: 'foo',
};

上面做了我们需要的；问题是当有几十个有效的属性名称时，它变得很麻烦。

Answer 1

您可以使用 distributive behavior of conditional types 从键的联合中自动创建联合：

type ValidRootPropertyNames =
'Property1' |
'Property2';

type ObjectWithRootPropertyNameFromList =
    ValidRootPropertyNames extends infer T ? // Introduce a type parameter to have distributive behavior 
    T extends string ? Record<T, string> : never : never;

const obj1: ObjectWithRootPropertyNameFromList = { // okay
    Property1: 'foo',
};

const obj2: ObjectWithRootPropertyNameFromList = { // okay
    Property2: 'foo',
};

const obj3: ObjectWithRootPropertyNameFromList = { // error
    Property3: 'foo',
};