javascript - 如何使用es6的filter或reduce来破坏对象属性？

Question

尝试使用reduce方法删除对象属性，但它没有返回预期的响应，在下面的用例中使用什么是正确的方法？过滤或减少？

main.js

const filtered = Object.keys(transformedResponse).reduce((res, key) => {

       delete res.drugName;
       delete res.mailPrice. copayEmployer
       delete res.retailPrice. copayEmployer
      return res;
    }, {});

转换后的响应

const transformedResponse = [
    {
        "isBrand": true,
        "drugName": "Lipitor",
        "drugStrength": "80 mg",
        "drugForm": "Tablet",
        "mailPrice": {
            "copayEmployer": 0,
            "prop2": "test"
        },
        "retialPrice": {
            "copayEmployer": 0,
            "prop2": "test"
        }

    }, {
        "isBrand": true,
        "drugName": "Metformin",
        "drugStrength": "500 mg",
        "drugForm": "Tablet",
        "mailPrice": {
            "copayEmployer": 50,
            "prop2": "test"
        },
        "retailPrice": {
            "copayEmployer": 0,
            "prop2": "test"
        }
    }

]

预期输出

[
    {
        "isBrand": true,
        "drugStrength": "80 mg",
        "drugForm": "Tablet",
        "mailPrice": {
            "prop2": "test"
        },
        "retialPrice": {
            "prop2": "test"
        }

    }, {
        "isBrand": true,
        "drugStrength": "500 mg",
        "drugForm": "Tablet",
        "mailPrice": {
            "prop2": "test"
        },
        "retailPrice": {
            "prop2": "test"
        }
    }

]

Answer 1

您可以使用map过滤结果

var x = transformedResponse.map((obj) => {
  return {
        "isBrand": obj.isBrand,
        "drugStrength": obj.drugStrength,
        "drugForm": obj.drugForm,
        "mailPrice": {
            "prop2": obj.mailPrice.prop2
        },
        "retailPrice": {
            "prop2": obj.retailPrice.prop2
        }
  }
});

console.log(x);

Map 迭代给定数组中的每个项目并返回一个新数组。 https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map