js 对象数组根据对象的key，进行筛选，合并？

Question

//原始数据
/**
  请你写一段代码
  将原始数据「filterList」与「selectedOptions」结合
  根据selectedOptions为true的项，产生与期望结果相符的结果
*/

const filterList = [
  {
    Code: "1",
    Id: "1",
    List: [
      { Code: '00', Name: '通用' }, { Code: '21', Name: '15秋' }, { Code: '22', Name: '15冬' },
      { Code: '23', Name: '16春' }, { Code: '24', Name: '16夏' }, { Code: '25', Name: '16秋' },
      { Code: '26', Name: '16冬' }, { Code: '27', Name: '17春' }, { Code: '28', Name: '17夏' },
      { Code: '29', Name: '17秋' }, { Code: '30', Name: '17冬' }, { Code: '31', Name: '18春' },
      { Code: '32', Name: '18夏' }, { Code: '33', Name: '18秋' }, { Code: '34', Name: '18冬' },
      { Code: '35', Name: '19春' }, { Code: '36', Name: '19夏' }, { Code: '37', Name: '19秋' },
      { Code: '38', Name: '19冬' }, { Code: '40', Name: '20夏' }, { Code: '41', Name: '20秋' },
      { Code: '45', Name: '20冬' }, { Code: '46', Name: '21秋' }, { Code: '47', Name: '21冬' },
      { Code: '48', Name: '21春' }, { Code: '49', Name: '21夏' }, { Code: '50', Name: '22春' },
      { Code: '51', Name: '22夏' }, { Code: '52', Name: '22冬' }, { Code: '53', Name: '22秋' },
      { Code: '54', Name: '2022春' }, { Code: '55', Name: '22秋冬' }
    ],
    ListType: "Base",
    Mode: "alone",
    Name: "销售季节",
    Type: "",
    inputType: "text",
  },
  {
    Code: "2",
    Id: "2",
    List: [
      { Code: 'DA', Name: '订货会订单' }, { Code: 'DH', Name: '期货订单' }, { Code: 'XS', Name: '档口订单' }
    ],
    ListType: "Base",
    Mode: "alone",
    Name: "单据类型",
    Type: "",
    inputType: "text",
  },
  {
    Code: "3",
    Id: "3",
    List: [
      { Code: '0', Name: '未审核' }, { Code: '1', Name: '已审核' }, { Code: '2', Name: '部分发' },
      { Code: '3', Name: '已发完' }, { Code: '4', Name: '已完成' }, { Code: '5', Name: '已作废' }
    ],
    ListType: "Base",
    Mode: "alone",
    Name: "单据状态",
    Type: "",
    inputType: "text",
  },
  {
    Code: "4",
    Id: "4",
    List: [
      { Code: '0', Name: 'OEM' }, { Code: '1', Name: 'TOB' }, { Code: '2', Name: '档口' },
      { Code: '3', Name: '订货会' }
    ],
    ListType: "Base",
    Mode: "alone",
    Name: "单据来源",
    Type: "",
    inputType: "text",
  },
  {
    Code: "5",
    Id: "5",
    List: [
      { Code: '01', Name: '订货' }, { Code: '02', Name: '订货买断' }, { Code: '03', Name: '订货追单' }
    ],
    ListType: "Base",
    Mode: "alone",
    Name: "订货类型",
    Type: "",
    inputType: "text",
  },
];

const selectedOptions = {
  "1_00": true,
  "1_49": false,
  "2_DH": true,
  "2_XS": true,
  "3_1": true,
  "4_1": true,
};

//期望结果
const expectedResult = [
  {
    Code: "1",
    ListType: "Base",
    Name: "销售季节",
    Type: "",
    Val: [
      { code: '00', val: '通用' }
    ],
    inputType: "text"
  },
  {
    Code: "2",
    ListType: "Base",
    Name: "单据类型",
    Type: "",
    Val: [
      { code: 'DH', val: '期货订单' },
      { code: 'XS', val: '档口订单' }
    ],
    inputType: "text"
  },
  {
    Code: "3",
    ListType: "Base",
    Name: "单据状态",
    Type: "",
    Val: [
      { code: '1', val: '已审核' }
    ],
    inputType: "text"
  },
  {
    Code: "4",
    ListType: "Base",
    Name: "单据来源",
    Type: "",
    Val: [
      { code: 'TOB', val: 'TOB' }
    ],
    inputType: "text"
  }
];

Answer 1

let list = filterList.map(({Id, Mode, List, ...v}) =>
          ({...v, List: List.filter(it => selectedOptions[`${v.Code}_${it.Code}`])}))
          .filter(({List}) => List.length)
console.log(list);

Answer 2

类似的答案已经解决了:
https://segmentfault.com/q/10...

Answer 3

跑是能跑，没考虑效率

let object1 = {};
const selectDNeedOptions = Object.keys(selectedOptions).map((key) => {
  const code = key.split('_')

  if (!object1.hasOwnProperty(code[0])) {
    const childobj = {}
    Object.defineProperty(childobj, code[1], {
      value: selectedOptions[key]
    })
    Object.defineProperty(object1, code[0], {
      value: childobj
    })

  } else {
    tmp = object1[code[0]]
    Object.defineProperty(tmp, code[1], {
      value: selectedOptions[key]
    })
  }


})

const result = filterList.filter((item) => {
  let curr = object1[item.Code]
  if (curr) {

    item.List = item.List.filter((a) => {
      if (curr[a.Code]) {

        return a
      }
    })
    return item
  }
})



console.log(result)

https://jsbin.com/repawoyihu/...

Answer 4

JavaScript 中的对象数组可以使用 Array.filter() 和 Array.reduce() 方法进行筛选和合并。

筛选：

const arr = [{id: 1, name: 'A'}, {id: 2, name: 'B'}, {id: 3, name: 'C'}];
const filteredArray = arr.filter(obj => obj.id === 2);

合并：

const arr = [{id: 1, value: 10}, {id: 2, value: 20}, {id: 3, value: 30}];
const mergedObj = arr.reduce((acc, obj) => {
  acc[obj.id] = obj.value;
  return acc;
}, {});

最后合并后的对象是： {1: 10, 2: 20, 3: 30}