通過js把一個數組修改成多層嵌套多個數組的幾種方法(轉)

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本文鏈接： https://blog.csdn.net/a13145211/article/details/104973859

1、首先一級與二級相同的嵌套

let arr = [
     { date: '2020-01-06', age: '18'},
     { date: '2020-01-06', age: '25'},
     { date: '2020-01-07', age: '34'},
     { date: '2020-01-07', age: '18'},
     { date: '2020-01-07', age: '38'},
     { date: '2020-01-08', age: '26'},
     { date: '2020-01-09', age: '24'}
   ]

let dataArr = [];
    arr.map(mapItem => {
      if (dataArr.length == 0) {
          dataArr.push({ date: mapItem.date, List: [mapItem] })
      } else {
         let res = dataArr.some(item=> {//判斷相同日期，有就添加到當前項
          if (item.date == mapItem.date) {
            item.List.push(mapItem)
            return true
          }
        })
        if (!res) {//如果沒找相同日期添加一個新對象
          dataArr.push({ date: mapItem.date, List: [mapItem] })
        }
      }
    })

結果

dataArr = [
        {
            date:'2020-01-06',
            List: [
                { date: '2020-01-06', age: '18'},
                { date: '2020-01-06', age: '25'}
            ]
        },
        {
            date:'2020-01-07',
            List: [
                { date: '2020-01-07', age: '34'},
                { date: '2020-01-07', age: '18'},
                { date: '2020-01-07', age: '38'},
            ]
        },
        {
            date:'2020-01-08',
            List: [
                 { date: '2020-01-08', age: '26'},
            ]
        },
        {
            date:'2020-01-09',
            List: [
                { date: '2020-01-09', age: '24' }
            ]
        }
    ]

第二種情況：一級四位數（0001），二級八位數字（00010001）獲取的條件是通過一級的四位數字匹配二級的前四位數字
方法1：

var userArr = [
  { id: "0001", userName: "laozhang_1" },
  { id: "00010001", userName: "laozhang1" },
  { id: "00010002", userName: "laozhang1" },
  { id: "00010003", userName: "laozhang1" },
  { id: "0002", userName: "laowang" },
  { id: "00020002", userName: "laozhang2" },
  { id: "00020003", userName: "laowang" },
  { id: "00020004", userName: "laowang" },
  { id: "00020005", userName: "laowang" },
  { id: "00020006", userName: "laowang" }
];

let newArr = []
 userArr.forEach(item => {
   if (item.id.length === 4) {
     newArr.push({ ...item, list: [] });
  }
 });
newArr.forEach(item => {
  userArr.forEach(user => {
     if (item.id !== user.id && item.id == user.id.substring(0, 4)) {
      item.list.push(user);
    }
  });
 });

console.log(newArr)

方法2：

var userArr = [
  { id: "0001", userName: "laozhang_1" },
  { id: "00010001", userName: "laozhang1" },
  { id: "00010002", userName: "laozhang1" },
  { id: "00010003", userName: "laozhang1" },
  { id: "0002", userName: "laowang" },
  { id: "00020002", userName: "laozhang2" },
  { id: "00020003", userName: "laowang" },
  { id: "00020004", userName: "laowang" },
  { id: "00020005", userName: "laowang" },
  { id: "00020006", userName: "laowang" }
];

let newArr = userArr.filter(item => item.id.length === 4)
newArr.forEach(item => {
      item.list = userArr.filter(user => item.id !== user.id && item.id == user.id.substring(0, 4))
    })
    console.log(newArr)

總結：學好js基礎很重要