NodeJS Json按子数组排序答案

【问题标题】：NodeJS Json sort by child arrayNodeJS Json按子数组排序
【发布时间】：2021-03-10 07:32:44
【问题描述】：

我想通过它的子数组对我的对象进行排序。

let partners = [
  {
    Name: "A",
    Sectors: [
      { Code: "AA1", Total: 10 },
      { Code: "AA2", Total: 70 }
    ]
  },
  {
    Name: "B",
    Sectors: [
      { Code: "AA1", Total: 20 },
      { Code: "AA2", Total: 60 }
    ]
  },
  {
    Name: "C",
    Sectors: [
      { Code: "AA1", Total: 40 },
      { Code: "AA2", Total: 50 }
    ]
  }
];

假设我想按Code = AA1 DESC 所在的支柱部门进行排序。结果（道具名称）将是：

C、B、A

或者如果我需要Code = AA2 DESC，将是 A、B、C

我遇到了一些像这样的简单排序的困难：

let sorted = partners.sort((a, b) => a.Sectors.Total - b.Sectors.Total);

sorted.forEach((sortedData) => console.log(sortedData));

Here is a SandBox

这里是正确的答案，将 acept empty 改为 acept empty。

let partners = [
{"Name":"E","Sectors": null}, 
{"Name":"A","Sectors":[{"Code":"AA1","Total":10},{"Code":"AA2","Total":70}]},
{"Name":"B","Sectors":[{"Code":"AA1","Total":20},{"Code":"AA2","Total":60}]},
{"Name":"C","Sectors":[{"Code":"AA1","Total":40},{"Code":"AA2","Total":50}]}, 
{"Name":"D","Sectors": null},
{"Name":"P","Sectors": []}
]

partners.sort((a, b) => {
    const get = (arr, key, val) => {
    return arr.find(o => o[key] === val) || {};
  }
  
  let sA = {}, sB = {}
  
  try {
    sA = get(a.Sectors, 'Code', 'AA1')
    sB = get(b.Sectors, 'Code', 'AA1')
  } catch (e) {}
  
  
  return !!sB.Total - !!sA.Total || sB.Total - sA.Total 
})

console.log(partners)

【问题讨论】：

标签： javascript node.js arrays json sorting

【解决方案1】：

您可以使用 sort 和 find 方法来做到这一点。在排序中，您需要从 Sectors 数组中找到具有特定 Code 的元素，然后比较 Total 的值。

let partners = [{"Name":"A","Sectors":[{"Code":"AA1","Total":10},{"Code":"AA2","Total":70}]},{"Name":"B","Sectors":[{"Code":"AA1","Total":20},{"Code":"AA2","Total":60}]},{"Name":"C","Sectors":[{"Code":"AA1","Total":40},{"Code":"AA2","Total":50}]}]

partners.sort((a, b) => {
  const sA = a.Sectors.find(({ Code }) => Code === 'AA1');
  const sB = b.Sectors.find(({ Code }) => Code === 'AA1');
  return sB.Total - sA.Total
})

console.log(partners)

【讨论】：

有时 a.Sectors 或 b.Sector 将为空或为空...空不是问题，我怎样才能做某种空值以使这种排序不会中断？
我在 NodeJS 12 上运行它，所以我没有问号 (?) 像 const sA = a?.Sectors?.find(({ Code }) => Code == = 'AA1');
你可以这样做jsfiddle.net/5bjuoLem。 !!sB.Total - !!sA.Total 这会将空值放在最后，sB.Total - sA.Total 在这里您可以修改 sB 和 sA 的顺序以获得 asc 或 desc 顺序。
不适用于empty´s...列表中有Sectors null 或Empty...
只是更改为返回 arr.find(o => o[key] === val) || {};并且工作得很好...但是 null 出现在 empty 前面...那条线...谢谢！

【解决方案2】：

您可以采用动态方法并找到所需的排序值。

const
    sortBy = (fn, order) => (a, b) => (order === 'ASC' || -1) * (fn(a) - fn(b)),
    getValue = code => o => o.Sectors.find(({ Code }) => Code === code)?.Total || 0,
    partners = [{ Name: "A", Sectors: [{ Code: "AA1", Total: 10 }, { Code: "AA2", Total: 70 }] }, { Name: "B", Sectors: [{ Code: "AA1", Total: 20 }, { Code: "AA2", Total: 60 }] }, { Name: "C", Sectors: [{ Code: "AA1", Total: 40 }, { Code: "AA2", Total: 50 }] }];

console.log(partners.sort(sortBy(getValue('AA1'), 'DESC')));
console.log(partners.sort(sortBy(getValue('AA2'), 'DESC')));

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【讨论】：