在Javascript

Question

如何在不使用任何库的情况下实现此目标？ 我已经尝试过使用一些ES6函数，但是最终复制了数组中的某些项目。它应该返回唯一，尤其是当数组中没有 child数组

时

我有三个数组变量：

data1 =第一个数据

data2 =要与 data1

合并的变量

data3 =合并变量的结果

        let data1 = [{
            "document_id": 12264,
            "detail_info": [{
                "id": 745,
                "lot_no": "X12345",
              },
              {
                "id": 744,
                "lot_no": "Z12345",
              }
            ]
          },
          {
            "document_id": 12226,
            "detail_info": [{
                "id": 738,
                "lot_no": "B12345",
              },
              {
                "id": 739,
                "lot_no": "C12345",
              }
            ]
          },
          {
            "document_id": 12221,
            "detail_info": []
          }
        ]

        let data2 = [{
            "document_id": 12264,
            "detail_info": [{
                "id": 744,
                "lot_no": "Z12345",
              },
              {
                "id": 743,
                "lot_no": "L12345",
              }
            ]
          },
          {
            "document_id": 12226,
            "detail_info": [{
              "id": 739,
              "lot_no": "C12345",
            }]
          },
          {
            "document_id": 12229,
            "detail_info": [{
              "id": 741,
              "lot_no": "E12345",
            }]
          },
          {
            "document_id": 10095,
            "detail_info": []
          }
        ]

//**This should be the result**
      let data3=[
      {
        "document_id": 12264,
        "detail_info": [
          {
            "id": 745,
            "lot_no": "X12345",
          },
          {
            "id": 744,
            "lot_no": "Z12345",
          },
          {
            "id": 743,
            "lot_no": "L12345",
          }
        ]
      },
      {
        "document_id": 12226,
        "detail_info": [
          {
            "id": 738,
            "lot_no": "B12345",
          },
          {
            "id": 739,
            "lot_no": "C12345",
          }
        ]
      },
      {
        "document_id": 12221,
        "detail_info": []
      },
      {
        "document_id": 12229,
        "detail_info": [
          {
            "id": 741,
            "lot_no": "E12345",
          }
        ]
      },
      {
        "document_id": 10095,
        "detail_info": []
      }
    ]

Answer 1

首先将data1和data2连接到单个数组（[...data1, ...data2]）中。

然后用Array.reduce()生成一个字典，该字典的键在数组中都可以找到document_id。

在reduce内，当document_id的条目已存在时，则必须将detail_info合并在一起。从现有条目和当前项创建所有detail_info的数组，并使用从Set生成的detail_info.id删除重复项，然后映射此Set id到相应的detail_info条目。

最后，使用Object.values()将字典转换为数组：

const data1 = [
  { "document_id": 12264, someData: 'hello', "detail_info": [{ "id": 745, "lot_no": "X12345" },  { "id": 744, "lot_no": "Z12345" }] },
  { "document_id": 12226, "detail_info": [{ "id": 738, "lot_no": "B12345" }, { "id": 739, "lot_no": "C12345" }] },
  { "document_id": 12221, "detail_info": [] }
];

const data2 = [
  { "document_id": 12264, "detail_info": [{ "id": 744, "lot_no": "Z12345" }, { "id": 743, "lot_no": "L12345" }] },
  { "document_id": 12226, "detail_info": [{ "id": 739, "lot_no": "C12345" }] },
  { "document_id": 12229, "detail_info": [{ "id": 741, "lot_no": "E12345" }] },
  { "document_id": 10095, "detail_info": [] }
];

const distinctById = arr => [...new Set(arr.map(({ id }) => id))]
  .map(id => arr.find(info => info.id === id))

const data3 = Object.values([...data1, ...data2].reduce((acc, x) => {
  acc[x.document_id] = acc[x.document_id] || { ...x, detail_info: [] };
  acc[x.document_id].detail_info = distinctById([...acc[x.document_id].detail_info, ...x.detail_info]);
  return acc;
}, {}));

console.log(data3);

如果要按插入顺序排列detail_info数组，请执行以下操作：

const distinctById = arr => {
  const uniqueIds = new Set(arr.map(({ id }) => id));
  return arr.filter(({ id }) => uniqueIds.delete(id));
}

console.log(distinctById([{ id: 5 }, { id: 4 }, { id: 5 }, { id: 3 }, { id: 4 }]))

Answer 2

如果只想合并两个数组，请使用concat：

let data3 = data1.concat(data2);