mongodb 中具有嵌套数组的多个文档分组

Question

我有一个文档列表，其中包含这些字段 ID：'int'、类别：'text'、字段：'text'，以及项目列表 [{title, value}]

     documents= [
      { "id":"1",
        "category": "education",
        "field": "science",
        "items": [
          {
            "title": "item 1",
            "value": "10"
          },
          {
            "title": "item 2",
            "value": "102"
          },
          {
            "title": "item 4",
            "value": "12"
          }
        ]
      },
      { "id":"2",
        "category": "education",
        "field": "science",
        "items": [
          {
            "title": "item 1",
            "value": "4"
          },
          {
            "title": "item 2",
            "value": "23"
          },
          {
            "title": "item 4",
            "value": "45"
          }
        ]
      },
      { "id":"3",
        "category": "fitness",
        "field": "body",
        "items": [
          {
            "title": "item 1",
            "value": "87"
          },
          {
            "title": "item 5",
            "value": "45"
          },
          {
            "title": "item =3",
            "value": "23"
          }
        ]
      },
 { "id":"4",
    "category": "education",
    "field": "culture",
    "items": [
      {
        "title": "item 1",
        "value": "187"
      },
      {
        "title": "item 5",
        "value": "145"
      },
      {
        "title": "item 3",
        "value": "123"
      }
    ]]

我正在与 mongodb（初学者）合作，我很困惑如何首先按类别对这些文档进行分组，然后按字段，然后按项目的标题将它们的值像列表一样推送到数组中价值观的历史。 期望的结果：

    newDocument=[
    { "newid":"1",
      "category": "education",
      "field": "science",
      "items": [
        {
          "title": "item 1",
          "value": ["10","4"]
        },
        {
          "title": "item 2",
          "value": ["102","23"]
        },
        {
          "title": "item 4",
          "value": ["12", "45"]
        }
      ]
    },
{ "newid":"2",
        "category": "education",
        "field": "culture",
        "items": [
            {
              "title": "item 1",
              "value": ["187"]
            },
            {
              "title": "item 5",
              "value":["145"]
            },
            {
              "title": "item 3",
              "value": ["123"]
            }
          ]
      }
    { "newid":"3",
        "category": "fitness",
        "field": "body",
        "items": [
            {
              "title": "item 1",
              "value": ["87"]
            },
            {
              "title": "item 5",
              "value":["45"]
            },
            {
              "title": "item 3",
              "value": ["23"]
            }
          ]
      }
  ]

Answer 1

可以使用聚合查询，但会影响性能和响应速度，有两种选择，一种使用$unwind，另一种使用$accumulator运算符，

使用 $accumulator:

• $group 通过 category 和 field
• $accumulator 做自定义 JavaScript 逻辑和合并项目数组，你可以根据你的要求改进和更新
• $project 显示必填字段

db.collection.aggregate([
  {
    $group: {
      _id: {
        category: "$category",
        field: "$field"
      },
      items: {
        $accumulator: {
          init: function() { return []; },
          accumulate: function(items1, items2) {
            return items1.concat(items2);
          },
          accumulateArgs: ["$items"],
          merge: function(items1, items2) {
            return items1.concat(items2);
          },
          finalize: function(state) { 
            var items = {}, finalItems = [];
            state.forEach(function(item) {
              if (items[item.title]) {
                items[item.title].value.push(item.value); 
              }
              else {
                item.value = [item.value];
                items[item.title] = item;
              }
            });
            for (var i in items) {
              finalItems.push(items[i]);
            }
            return finalItems; 
          },
          lang: "js"
        }
      }
    }
  },
  {
    $project: {
      _id: 0,
      category: "$_id.category",
      field: "$_id.field",
      items: 1
    }
  }
])

使用 $unwind：（不推荐）

• $unwind解构items数组
• $group by category, field and item title 并构造数组 if item values
• $group 通过 category 和 field 并用 title 和 value
构造 items 数组
• $project 显示必填字段

db.collection.aggregate([
  { $unwind: "$items" },
  {
    $group: {
      _id: {
        category: "$category",
        field: "$field",
        title: "$items.title"
      },
      value: { $push: "$items.value" }
    }
  },
  {
    $group: {
      _id: {
        category: "$_id.category",
        field: "$_id.field"
      },
      items: {
        $push: {
          title: "$_id.title",
          value: "$value"
        }
      }
    }
  },
  {
    $project: {
      _id: 0,
      category: "$_id.category",
      field: "$_id.field",
      items: 1
    }
  }
])

Playground