Typescript 更新ImmutableJS记录的多个字段的推荐方法是什么？

我有ImmutableJS记录，我想更新多个字段，推荐的方法是什么

我尝试过变异，但由于某种原因，我得到的副本是空的，我无法对其进行变异：

const UserRecord = Record({
  firstName: null,
  lastName: null, 
  address: {
    city: null, 
    zipCode: null
  }
});

export function createUserRecord(data) {
  return new UserRecord(fromJS(data));
}

let u = createUserRecord({
  firstName: 'John',
  lastName: 'Don', 
  address: {
    city: 'New York', 
    zipCode: 1111
  }
});

let newU = u.withMutations((uCopy) => {
  // uCopy is empty (with default values)
  uCopy.set('firstName', 'New FirstName');
  uCopy.set('lastName', 'New LastName');
  // uCopy is still empty (not mutated)
});

console.log(u.toJS()); // {firstName: "John", lastName: "Don", address: {city: null, zipCode: null}}
console.log(newU.toJS()); // {firstName: null, lastName: null, address: {city: null, zipCode: null}}

---

谢谢大家!

带有突变的方法

不会改变对象
u
。它返回一个变异的对象，所以您需要像这样调用它

let newU = u.withMutations((uCopy) => {
  // uCopy is empty (with default values)
  uCopy.set('firstName', 'New FirstName');
  uCopy.set('lastName', 'New LastName');
  // uCopy is still empty (not mutated)
});


请看来自的文档中的这个示例

我已编辑了问题，请再次检查。谢谢你的回答！我将您的代码复制到JSFIDLE。结果是-
Object{firstName:“John”，lastName:“Don”，address:Object}Object{firstName:“newfirstname”，lastName:“New lastName”，address:Object}
。对象
u
保持原样，您得到的新对象，即
newU
具有firstName和lastName的新值，街道仍然是纽约1111
var list1 = Immutable.List.of(1,2,3);
var list2 = list1.withMutations(function (list) {
  list.push(4).push(5).push(6);
});
assert(list1.size === 3);
assert(list2.size === 6);