Question

我有多个文档，我正在尝试汇总companyId = xxx的所有文档，并返回一个包含所有状态的数组。

所以它看起来像这样：

[
{
    "status": "created",
    "date": "2019-03-16T10:59:59.200Z"
},
{
    "status": "completed",
    "date": "2019-03-16T11:00:37.750Z"
},
{
    "status": "created",
    "date": "2019-03-16T10:59:59.200Z"
},
{
    "status": "completed",
    "date": "2019-03-16T11:00:37.750Z"
},
{
    "status": "created",
    "date": "2019-03-16T10:59:59.200Z"
},
{
    "status": "completed",
    "date": "2019-03-16T11:00:37.750Z"
},
{
    "status": "created",
    "date": "2019-03-16T10:59:59.200Z"
},
{
    "status": "completed",
    "date": "2019-03-16T11:00:37.750Z"
}

]

文档如下：

[
{
    "companyId": "xxx",
    "position": "",
    "section": "",
    "comment": "",
    "items": [
        {
            "any": "111",
            "name": "some name",
            "description": "some description",
            "version": "3",
            "status": [
                {
                    "status": "created",
                    "date": "2019-03-16T10:59:59.200Z"
                },
                {
                    "status": "completed",
                    "date": "2019-03-16T11:00:37.750Z"
                }
            ]
        },
        {
            "any": "222",
            "name": "some name",
            "description": "some description",
            "version": "3",
            "status": [
                {
                    "status": "created",
                    "date": "2019-03-16T10:59:59.200Z"
                },
                {
                    "status": "completed",
                    "date": "2019-03-16T11:00:37.750Z"
                }
            ]
        }
    ]
},
{
    "companyId": "xxx",
    "position": "",
    "section": "",
    "comment": "",
    "items": [
        {
            "any": "111",
            "name": "some name",
            "description": "some description",
            "version": "3",
            "status": [
                {
                    "status": "created",
                    "date": "2019-03-16T10:59:59.200Z"
                },
                {
                    "status": "completed",
                    "date": "2019-03-16T11:00:37.750Z"
                }
            ]
        },
        {
            "any": "222",
            "name": "some name",
            "description": "some description",
            "version": "3",
            "status": [
                {
                    "status": "created",
                    "date": "2019-03-16T10:59:59.200Z"
                },
                {
                    "status": "completed",
                    "date": "2019-03-16T11:00:37.750Z"
                }
            ]
        }
    ]
}

]

任何建议，如何实施？

然后，我想遍历数组（在代码中）并计算状态中创建和完成的项目数。也许可以通过查询完成？

预先感谢

Answer 1

除了@mickl答案外，您还可以添加$project管道以将结果作为状态和计数的平面列表来获取。

db.collectionName.aggregate([
    {
        $match: { companyId: "xxx" }
    },
    {
        $unwind: "$items"
    },
    {
        $unwind: "$items.status"
    },
    {
        $replaceRoot: {
            newRoot: "$items.status"
        }
    },
    {
        $group: {
            _id: "$status",
            count: { $sum: 1 }
        }
    },
    {
        $project: {
            "status":"$_id", 
            "count":1,
            _id:0
        }
    }
])

Answer 2

您可以使用以下汇总：

db.col.aggregate([
    {
        $match: { companyId: "xxx" }
    },
    {
        $unwind: "$items"
    },
    {
        $unwind: "$items.status"
    },
    {
        $replaceRoot: {
            newRoot: "$items.status"
        }
    },
    {
        $group: {
            _id: "$status",
            count: { $sum: 1 }
        }
    }
])

双$unwind将为每个文档返回一个状态，然后您可以使用$replaceRoot将每个状态提升到文档的根目录级别。

此外，您可以添加$group阶段以按status计数文档。

Answer 3

如果要执行上述查询的文档数量过多，则应避免在聚合管道的初始阶段使用 $ unwind 。
您应该在 $ match 之后使用 $ project 来减少字段的选择，或者可以使用以下查询：

db.col.aggregate([
  {
    $match: {
      companyId: "xxx"
    }
  },
  {
    $project: {
      _id: 0,
      data: {
        $reduce: {
          input: "$items.status",
          initialValue: [

          ],
          in: {
            $concatArrays: [
              "$$this",
              "$$value"
            ]
          }
        }
      }
    }
  },
  {
    $unwind: "$data"
  },
  {
    $replaceRoot: {
      newRoot: "$data"
    }
  }
])

聚合嵌套数组

3 个答案: