Typescript 在不使用任何对象的情况下访问对象的子对象

使用当前参数类型setValue，可以将child类B提供为fieldType(其中extends user[fieldName]为A类)，并以B的fieldAttribute为目标，这可能不存在于A中:

class PhoneNumber2 extends PhoneNumber {
  extension: number;
}

// Complies with setValue types,
// but if `user` already has a basic PhoneNumber instance,
// the code within setValue fails
setValue(408, PhoneNumber2, "phone", "extension");

// within setValue:
user["phone"]["extension"] = 408;
// ...but user["phone"] may be a PhoneNumber,
// not necessarily a PhoneNumber2!

在您的例子中，不是K extends keyof T(T是fieldType的返回类型，可能是子类)，而是只需使用K extends keyof User[A]，以确保fieldAttribute指向基类:

function setValue2<A extends keyof User, T extends User[A], K extends keyof User[A], V extends T[K]>(
  value: V,
  fieldType: new () => T,
  fieldName: A,
  fieldAttribute: K
) {

  if (!user[fieldName]) {
    user[fieldName] = new fieldType();
  }

  const field = user[fieldName];

  field[fieldAttribute] = value; // Now okay
}

setValue2(408, PhoneNumber, 'phone', 'prefix');
setValue2(408, PhoneNumber2, 'phone', 'prefix');
setValue2(408, PhoneNumber2, 'phone', 'extension'); // Error: Argument of type '"extension"' is not assignable to parameter of type 'keyof PhoneNumber'.

您甚至可以go 掉泛型类型T:

function setValue3<A extends keyof User, K extends keyof User[A], V extends User[A][K]>(
    value: V,
    fieldType: new () => User[A],
    fieldName: A,
    fieldAttribute: K
) {}

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