聚合嵌套数组

Israel 发表于 Dev

以色列

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

因此它将如下所示：

[
{
    "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"
}

]

该文档如下所示：

[
{
    "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"
                }
            ]
        }
    ]
}

]

有什么建议，如何实施？

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

提前致谢

米克尔

您可以使用以下聚合：

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

双重$ unwind将为每个文档返回单个状态，然后您可以使用$ replaceRoot将每个状态提升为文档的根级别。

另外，您可以添加$ group阶段以按计数文档status。

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