js实现数组根据相同字段，合并出新的数据？

• Z时代
• 2024-03-16
• 分类：IT

把相同batch合并，并生成对应的positionList 数组
原始数组：

    const list = [
      {
        positionId: '1001',
        positionName: 'B-b-b-101',
        batch: '1',
        num: 5,
      },
      {
        positionId: '1002',
        positionName: 'B-b-b-102',
        batch: '2',
        num: 3,
      },
      {
        positionId: '1004',
        positionName: 'B-b-b-104',
        batch: '1',
        num: 1,
      },
      {
        positionId: '1004',
        positionName: 'B-b-b-105',
        batch: '2',
        num: 3,
      },
    ]

期望数组：

  const list = [
      {
        positionId: '1001',
        positionName: 'B-b-b-101',
        batch: '1',
        num: 5,
        positionList: [
          {
            positionId: '1001',
            positionName: 'B-b-b-101',
            num: 5,
          },
          {
            positionId: '1004',
            positionName: 'B-b-b-104',
            num: 1,
          },
        ],
      },
      {
        positionId: '1002',
        positionName: 'B-b-b-102',
        batch: '2',
        num: 3,
        positionList: [
          {
            positionId: '1002',
            positionName: 'B-b-b-102',
            num: 2,
          },
        ],
      },
      {
        positionId: '1005',
        positionName: 'B-b-b-105',
        batch: '2',
        num: 3,
        positionList: [
          {
            positionId: '1005',
            positionName: 'B-b-b-105',
            num: 3,
          },
        ],
      },
    ]

回答：

    const list = [
      {
        positionId: '1001',
        positionName: 'B-b-b-101',
        batch: '1',
        num: 5,
      },
      {
        positionId: '1002',
        positionName: 'B-b-b-102',
        batch: '2',
        num: 3,
      },
      {
        positionId: '1004',
        positionName: 'B-b-b-104',
        batch: '1',
        num: 1,
      },
      {
        positionId: '1004',
        positionName: 'B-b-b-105',
        batch: '2',
        num: 3,
      },
    ]
let obj ={}
let newList = list.reduce((item, next) => {
    obj[next.batch] ? '' : obj[next.batch] = true && item.push(next)
    return item
}, []).map(m=>{
    m.positionList = JSON.parse(JSON.stringify(list.filter(f => f.batch == m.batch)))
    return m
})
console.log(newList)//格式化好的数据

回答：

js 代码

Object.values(list.reduce((obj, {batch, ...rest}) => {
    obj[batch] ??= {batch, ...rest, positionList: []};
    obj[batch].positionList.push({...rest});
    return obj;
}, {}))

结果

[
  {
    "batch": "1",
    "positionId": "1001",
    "positionName": "B-b-b-101",
    "num": 5,
    "positionList": [
      {"positionId": "1001", "positionName": "B-b-b-101", "num": 5},
      {"positionId": "1004", "positionName": "B-b-b-104", "num": 1}
    ]
  },
  {
    "batch": "2",
    "positionId": "1002",
    "positionName": "B-b-b-102",
    "num": 3,
    "positionList": [
      {"positionId": "1002", "positionName": "B-b-b-102", "num": 3},
      {"positionId": "1004", "positionName": "B-b-b-105", "num": 3}
    ]
  }
]

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