javascript - 如何使用 JS spread rest destruction 删除嵌套属性？

Question

尝试删除多个属性 usign spread operator rest destruction，所以 drugName 被删除但我有嵌套对象 mailPrice.copayEmployer 没有被破坏不知道什么是实现错误？

主要.js

const transformedResponse = transformResponse(response);
  const loggerResponse = transformedResponse.map(
      ({drugName, mailPrice.copayEmployer, ...rest}) => rest
  );

转换响应

[{
        "isBrand": true,
        "drugName": "Lipitor",
        "drugStrength": "80 mg",
        "drugForm": "Tablet",
        "mailPrice": {
            "copayEmployer": 0
        }
    }, {
        "isBrand": true,
        "drugName": "Metformin",
        "drugStrength": "500 mg",
        "drugForm": "Tablet",
        "mailPrice": {
            "copayEmployer": 50
        }
    }

]

Answer 1

我不认为你可以用一个变量来做到这一点。您需要在 mailPrice 对象中嵌套一个 ...rest 变量，然后您必须重建该对象。

const transformedResponse = [{
  "isBrand": true,
  "drugName": "Lipitor",
  "drugStrength": "80 mg",
  "drugForm": "Tablet",
  "mailPrice": {
    "copayEmployer": 0,
  }
}, {
  "isBrand": true,
  "drugName": "Metformin",
  "drugStrength": "500 mg",
  "drugForm": "Tablet",
  "mailPrice": {
    "copayEmployer": 50
  }
}];

const loggerResponse = transformedResponse.map(
  ({
    drugName,
    mailPrice: { copayEmployer, ...mailPriceRest},
    ...rest
  }) => ({mailPrice: {...mailPriceRest}, ...rest})
);

console.log(loggerResponse);