如何在Javascript/Typescript中正确使用spread operator[duplicate]

Question

我有两个不同类型的对象，我想用一个对象指定另一个。

interface From {
  type: string;
  timestamp: number;
  aid?: string;
  bid?: string;
  cid?: string;
  did?: string;
  eid?: string;
  fid?: string;
  gid?: string;
}

interface To {
  fromSocketID: string;
  type: string;
  timestamp: number;
  aid?: string;
  bid?: string;
  cid?: string;
  did?: string;
  eid?: string;
  fid?: string;
}

const from: From = {
  type: "aaa",
  timestamp: 1231231231,
  gid: "ddd"
};

// Method1
const to1: To = {
  fromSocketID: "temp",
  type: from.type,
  timestamp: from.timestamp
};
if (from.aid) {
  to1.aid = from.aid
}
if (from.bid) {
  to1.bid = from.bid;
}
if (from.cid) {
  to1.cid = from.cid;
}
// ...three more if statements

// Method2
const to2: To = {
  fromSocketID: "temp",
  ...from
}
// @ts-ignore
delete to2.gid;

到的接口有一个fromSocketID，而从到没有gid属性。在我的实际工作场景中，我使用Method1。我尝试了Method2，但我不得不使用＜code＞ts忽略＜/code＞。我想知道是否有更好的解决办法。




Answer 1

您可以使用 rest 运算符解构 'from'，忽略 gid 属性，如下所示：

interface From {
  type: string;
  timestamp: number;
  aid?: string;
  bid?: string;
  cid?: string;
  did?: string;
  eid?: string;
  fid?: string;
  gid?: string;
}

interface To {
  fromSocketID: string;
  type: string;
  timestamp: number;
  aid?: string;
  bid?: string;
  cid?: string;
  did?: string;
  eid?: string;
  fid?: string;
}

const from: From = {
  type: "aaa",
  timestamp: 1231231231,
  gid: "ddd"
};

const { gid, ...rest } = from;

const to: To = {
  fromSocketID: "temp",
  ...rest
};