javascript转换成父子数组

Question

任何人都可以帮助转换以下父子对象列表：

我下面有对象数组，需要将其转换为父子顺序。 对象中的每个“成员”属性中可能包含1个或n个对象。 在“成员”数组中，第一个对象是第二个对象的父对象，第二个对象是第三个对象的父对象。

因此，在第1个成员中，“视频”是“西方”的父母，“西方”是“印度”的父母，依此类推。

我试图一遍一遍地遍历元素，但无法达到预期的结果。

逻辑或代码方面的任何帮助将非常有帮助。

输入：

[
  {
    "Members": [
      {
        "Name": "Videos"
      },
      {
        "Name": "West"
      },
      {
        "Name": "India"
      }
    ]
  },
  {
    "Members": [
      {
        "Name": "Machinery"
      },
      {
        "Name": "South"
      },
      {
        "Name": "Australia"
      }
    ]
  },
  {
    "Members": [
      {
        "Name": "Electronics"
      },
      {
        "Name": "Midwest"
      },
      {
        "Name": "Arab"
      }
    ]
  },
  {
    "Members": [
      {
        "Name": "Machinery"
      },
      {
        "Name": "West"
      },
      {
        "Name": "India"
      }
    ]
  },
  {
    "Members": [
      {
        "Name": "Electronics"
      },
      {
        "Name": "NorthEast"
      },
      {
        "Name": "Japan"
      }
    ]
  },
  {
    "Members": [
      {
        "Name": "Videos"
      },
      {
        "Name": "South"
      },
      {
        "Name": "Australia"
      }
    ]
  },
  {
    "Members": [
      {
        "Name": "Videos"
      },
      {
        "Name": "West"
      },
      {
        "Name": "Japan"
      }
    ]
  }
]

预期产量：

[
  {
    "name": "Videos",
    "children": [
      {
        "name": "West",
        "children": [
          {
            "name": "India",
            "children": []
          },
          {
            "name": "Japan",
            "children": []
          }
        ]
      },
      {
        "name": "South",
        "children": [
          {
            "name": "Australia",
            "children": []
          }
        ]
      }
    ]
  },
  {
    "name": "Machinery",
    "children": [
      {
        "name": "South",
        "children": [
          {
            "name": "Australia",
            "children": []
          }
        ]
      },
      {
        "name": "West",
        "children": [
          {
            "name": "India",
            "children": []
          }
        ]
      }
    ]
  },
  {
    "name": "Electronics",
    "children": [
      {
        "name": "Midwest",
        "children": [
          {
            "name": "Arab",
            "children": []
          }
        ]
      },
      {
        "name": "NorthEast",
        "children": [
          {
            "name": "Japan",
            "children": []
          }
        ]
      }
    ]
  }
]

```

Answer 1

男人这花了太长时间。 但它适用于更大的数据集。 注意OP，切勿使用此数据结构。 曾经 这太糟糕了。 我失去了许多解决此问题的方法：

 var arr = [ {Members: [{ Name: "Videos" }, { Name: "West" }, { Name: "India" }, {Name: 'Testing'}]}, {Members: [{ Name: "Machinery" }, { Name: "South" }, { Name: "Australia" }]}, {Members: [{ Name: "Electronics" }, { Name: "Midwest" }, { Name: "Arab" }]}, {Members: [{ Name: "Machinery" }, { Name: "West" }, { Name: "India" }]}, {Members: [{ Name: "Electronics" }, { Name: "NorthEast" }, { Name: "Japan" }]}, {Members: [{ Name: "Videos" }, { Name: "South" }, { Name: "Australia" }]}, {Members: [{ Name: "Videos" }, { Name: "West" }, { Name: "Japan" }]} ]; const addRelation = (obj, m, i) => ({...obj, parent: i === 0 ? null : m.slice(0, i).map(el => el.Name).join('.'), level: i, children: []}) const arrayToTree = (arr) => { arr = arr.map(({ Members: m }) => m.map((obj, i) => addRelation(obj, m, i))).reduce((acc, arr) => { arr.map(obj => acc.push(obj)) return acc }, []).sort((a, b) => b.level - a.level) var temp = [...arr].filter((o, index, self) => index === self.findIndex((t) => ( t.Name === o.Name && t.parent === o.parent )) ) arr.forEach(() => { if (temp[0].level === 0) return var parentIndex = temp.findIndex(o => { var str = temp[0].parent var rest = str.substring(0, str.lastIndexOf(".")); var last = str.substring(str.lastIndexOf(".") + 1, str.length); var parents = [rest, last] return parents[0] !== '' ? o.Name === parents[1] && o.parent === parents[0] : o.Name === temp[0].parent }) const { Name, children } = temp[0] temp[parentIndex].children.push({Name, children}) temp.shift() }) return temp.map(({ Name, children }) => ({ Name, children })) } arr = arrayToTree(arr) console.log(arr) 

Answer 2

这可能不是解决此问题的最佳方法，并且仅适用于3个级别。

 var data = [ { "Members": [ { "Name": "Videos" }, { "Name": "West" }, { "Name": "India" } ] }, { "Members": [ { "Name": "Machinery" }, { "Name": "South" }, { "Name": "Australia" } ] }, { "Members": [ { "Name": "Electronics" }, { "Name": "Midwest" }, { "Name": "Arab" } ] }, { "Members": [ { "Name": "Machinery" }, { "Name": "West" }, { "Name": "India" } ] }, { "Members": [ { "Name": "Electronics" }, { "Name": "NorthEast" }, { "Name": "Japan" } ] }, { "Members": [ { "Name": "Videos" }, { "Name": "South" }, { "Name": "Australia" } ] }, { "Members": [ { "Name": "Videos" }, { "Name": "West" }, { "Name": "Japan" } ] } ]; function organize(dataBefore){ var dataAfter = []; dataAfter.push({ name: dataBefore[0].Members[0].Name, children: [] }); for(var i = 1; i < dataBefore.length; i++){ if(!doesExist(dataAfter, dataBefore[i].Members[0].Name)){ dataAfter.push({ name: dataBefore[i].Members[0].Name, children: [] }); } } dataAfter[0].children.push({ name: dataBefore[0].Members[1].Name, children: [] }); for(var i = 1; i < dataBefore.length; i++){ for(var j = 0; j < dataAfter.length; j++){ if(dataAfter[j].name == dataBefore[i].Members[0].Name){ if(!doesExist(dataAfter[j].children, dataBefore[i].Members[1].Name)){ dataAfter[j].children.push({ name: dataBefore[i].Members[1].Name, children: [] }); } } } } dataAfter[0].children[0].children.push({ name: dataBefore[0].Members[2].Name, children: [] }); for(var i = 1; i < dataBefore.length; i++){ for(var j = 0; j < dataAfter.length; j++){ if(dataAfter[j].name == dataBefore[i].Members[0].Name){ for(var k = 0; k < dataAfter[j].children.length; k++){ if(dataAfter[j].children[k].name == dataBefore[i].Members[1].Name){ if(!doesExist(dataAfter[j].children[k].children, dataBefore[i].Members[2].Name)){ dataAfter[j].children[k].children.push({ name: dataBefore[i].Members[2].Name, children: [] }); } } } } } } return dataAfter; } function doesExist(checkThisData, searchValue){ for(var i = 0; i < checkThisData.length; i++){ if(searchValue == checkThisData[i].name){ return true; } } return false; } console.log(organize(data)); 

Answer 3

我建议使用map + reduce组合：

 const data = [{ "Members": [{ "Name": "Videos" }, { "Name": "West" }, { "Name": "India" } ] }, { "Members": [{ "Name": "Machinery" }, { "Name": "South" }, { "Name": "Australia" } ] }, { "Members": [{ "Name": "Electronics" }, { "Name": "Midwest" }, { "Name": "Arab" } ] }, { "Members": [{ "Name": "Machinery" }, { "Name": "West" }, { "Name": "India" } ] }, { "Members": [{ "Name": "Electronics" }, { "Name": "NorthEast" }, { "Name": "Japan" } ] }, { "Members": [{ "Name": "Videos" }, { "Name": "South" }, { "Name": "Australia" } ] }, { "Members": [{ "Name": "Videos" }, { "Name": "West" }, { "Name": "Japan" } ] } ]; const ar = data.map((el, i) => { let a = el['Members'].reduce((acc, member, index) => { if (index === 0) { acc[index] = { name: member.Name, children: [] }; } else { debugger; if (acc[0].children.length === 0) { acc[0].children.push({ name: member.Name, children: [] }); } else { acc[0].children[0].children.push({ name: member.Name, children: [] }); } } return acc; }, []); return a; }); console.log(ar);