如何将数据设置为深度嵌套数组

Question

我正在使用 DevExtreme React 网格树。我的初始调用仅填充根行，单击时应用每个附加子行。当有许多嵌套表行时，我在应用子行数据时遇到问题。我需要一种有效的方法来找到正确的父行并添加下一个嵌套数组。这是我已经添加的带有一个嵌套行的表数据。

    [
  {
    "area": "Artesia",
    "list_id": 45,
    "rowId": 158324175700860960,
    "parentRowId": 0,
    "items": [
      {
        "area": "Other",
        "list_id": 15003,
        "rowId": 158324179061139520,
        "parentRowId": 158324175700860960
      }
    ]
  },
  {
    "area": "Corpus Christi",
    "list_id": 60,
    "rowId": 158324175700847800,
    "parentRowId": 0,
    "items": []
  },
  {
    "area": "Midland",
    "list_id": 10,
    "rowId": 158324175700867700,
    "parentRowId": 0,
    "items": [
      {
        "area": "Delaware Basin",
        "list_id": 11001,
        "rowId": 158324181266273440,
        "parentRowId": 158324175700867700,
        "items": []
      }
    ]
  },
  {
    "area": "San Antonio",
    "list_id": 63,
    "rowId": 158324175700814050,
    "parentRowId": 0,
    "items": []
  }
]

单击 Midland 行后，我将 API 返回数据应用为嵌套数组项。

    [
  {
    "area": "Delaware Basin",
    "list_id": 11001,
    "rowId": 158324181266273440,
    "parentRowId": 158324175700867700,
    "items": []
  }
]

我需要一个函数，它可以将表数据从根级别循环到无限的嵌套行。我现在通过使用 parentId 来匹配 rowId 来匹配数据。

    [
  {
    "area": "Artesia",
    "list_id": 45,
    "rowId": 158324175700860960,
    "parentRowId": 0,
    "items": [
      {
        "area": "Other",
        "list_id": 15003,
        "rowId": 158324179061139520,
        "parentRowId": 158324175700860960
      }
    ]
  },
  {
    "area": "Corpus Christi",
    "list_id": 60,
    "rowId": 158324175700847800,
    "parentRowId": 0,
    "items": []
  },
  {
    "area": "Midland",
    "list_id": 10,
    "rowId": 158324175700867700,
    "parentRowId": 0,
    "items": [
      {
        "area": "Delaware Basin",
        "list_id": 11001,
        "rowId": 158324181266273440,
        "parentRowId": 158324175700867700,
        "items": []
      }
    ]
  },
  {
    "area": "San Antonio",
    "list_id": 63,
    "rowId": 158324175700814050,
    "parentRowId": 0,
    "items": []
  }
]

Answer 1

您可以使用一些 dfs 算法简单地遍历树

const tableData = [{ area: 'Artesia ', list_id: 45, rowId: 15836049402342958, parentRowId: 0, Premium: '785', 'Non Premium': '152', Total: '937', items: [] }, { area: 'Corpus Christi ', list_id: 60, rowId: 158360494023429300, parentRowId: 0, Total: '3,098', items: [] }, { area: 'Denver ', list_id: 30, rowId: 158360494023563870, parentRowId: 0, Total: '7,938', items: [] }, { area: 'Fort Worth ', list_id: 14, rowId: 158360494023592130, parentRowId: 0, Total: '14', items: [{ area: 'Southlake', list_id: 1256788, rowId: 12345, parentRowId: 158360494023592130, Premium: '7,743', 'Non Premium': '2,584', Total: '10,327', items: [] }] }, { area: 'Midland ', list_id: 10, rowId: 158360494023510200, parentRowId: 0, Premium: '7,743', 'Non Premium': '2,584', Total: '10,327', items: [{ area: 'Delaware Basin', list_id: 11001, rowId: 158324181266273440, parentRowId: 158360494023510200, Premium: '7,743', 'Non Premium': '2,584', Total: '10,327', items: [] }] }, { area: 'Okc ', list_id: 50, rowId: 158360494023542430, parentRowId: 0, Total: '245', items: [] }, { area: 'San Antonio ', list_id: 63, rowId: 158360494023591680, parentRowId: 0, Total: '4,162', items: [] }]
const returnData = [{ area: 'Delaware Basin Nm', list_id: 11007, rowId: 158324182577224580, parentRowId: 158324181266273440 }, { area: 'Delaware Basin Tx', list_id: 11002, rowId: 158324182577248960, parentRowId: 158324181266273440 }, { area: 'Sub Southlake', list_id: 2345, rowId: 550987654, parentRowId: 12345 }]

const byParentRowId = returnData.reduce((m, it) => {
  const v = m.get(it.parentRowId) || []
  v.push(it)
  m.set(it.parentRowId, v)
  return m
}, new Map())


const findRow = (tableData => {
  function dfs (data, stopId) {
    if (data.rowId === stopId) return data
    if (!Array.isArray(data.items)) return []
    return data.items.flatMap(x => dfs(x, stopId))
  }
  return rowId => dfs({ items: tableData }, rowId)[0]
})(tableData)

console.time('setTree1')
;[...byParentRowId.entries()].forEach(([rowId, v]) => (findRow(rowId).items = v))
console.timeEnd('setTree1')
console.log(JSON.stringify({ items: tableData }, null, 2))

请注意，对于每个不同的 parentRowId，您都要遍历树。如果您想变得更加精明，但需要付出更多代码，您可以预先构建一个映射 rowId => 节点，并在填充嵌套行时对其进行维护。我建议不要这样做，除非您观察到显着（且有意义）的收益。在这里，它是1ms，所以没用。

const tableData = [{ area: 'Artesia ', list_id: 45, rowId: 15836049402342958, parentRowId: 0, Premium: '785', 'Non Premium': '152', Total: '937', items: [] }, { area: 'Corpus Christi ', list_id: 60, rowId: 158360494023429300, parentRowId: 0, Total: '3,098', items: [] }, { area: 'Denver ', list_id: 30, rowId: 158360494023563870, parentRowId: 0, Total: '7,938', items: [] }, { area: 'Fort Worth ', list_id: 14, rowId: 158360494023592130, parentRowId: 0, Total: '14', items: [{ area: 'Southlake', list_id: 1256788, rowId: 12345, parentRowId: 158360494023592130, Premium: '7,743', 'Non Premium': '2,584', Total: '10,327', items: [] }] }, { area: 'Midland ', list_id: 10, rowId: 158360494023510200, parentRowId: 0, Premium: '7,743', 'Non Premium': '2,584', Total: '10,327', items: [{ area: 'Delaware Basin', list_id: 11001, rowId: 158324181266273440, parentRowId: 158360494023510200, Premium: '7,743', 'Non Premium': '2,584', Total: '10,327', items: [] }] }, { area: 'Okc ', list_id: 50, rowId: 158360494023542430, parentRowId: 0, Total: '245', items: [] }, { area: 'San Antonio ', list_id: 63, rowId: 158360494023591680, parentRowId: 0, Total: '4,162', items: [] }]
const returnData = [{ area: 'Delaware Basin Nm', list_id: 11007, rowId: 158324182577224580, parentRowId: 158324181266273440 }, { area: 'Delaware Basin Tx', list_id: 11002, rowId: 158324182577248960, parentRowId: 158324181266273440 }, { area: 'Sub Southlake', list_id: 2345, rowId: 550987654, parentRowId: 12345 }]

const byParentRowId = returnData.reduce((m, it) => {
  const v = m.get(it.parentRowId) || []
  v.push(it)
  m.set(it.parentRowId, v)
  return m
}, new Map())

const table = (tableData => {
  const rows = new Map()
  function dfs (data) {
    if (data.rowId) {
      rows.set(data.rowId, data)
    }
    if (!Array.isArray(data.items)) { return }
    return data.items.forEach(dfs)
  }
  dfs({ items: tableData })
  return {
    setRow: (rowId, items) => {
      items.forEach(it => rows.set(it.rowId, it))
      const row = rows.get(rowId)
      row.items = items
    },
    getRow: rowId => rows.get(rowId)
  }
})(tableData)

console.time('setTree2')
;[...byParentRowId.entries()].forEach(([rowId, v]) => table.setRow(rowId, v))
console.timeEnd('setTree2')
console.log(JSON.stringify({items: tableData},null,2))