如何使用聚合 MongoDB 3.0 版获取 Mongo 数据

Question

假设我们有 10 个集合，那么我们必须根据 tag_id 找到计数。例如，如果tag_id包含0和1，那么我们必须统计所有数据，以及统计没有tag_id的数据，或者tag_id为null的数据。然后如果它有 unread : false 那么输出来，所有未读的计数。

查找 tag_id 的计数和 false 时的未读计数。

{
        "_id": ObjectId("5912c7240520df77f0c2c18a"),
        "email_id": "54",
        "unread": "false",
        "__v": NumberLong(0),
        "tag_id": ["0"

        ]
    }, {
        "_id": ObjectId("5912c71e0520df77f0c2c189"),
        "email_id": "55",
        "unread": "false",
        "__v": NumberLong(0),
        "tag_id": [
            "1"
        ]
    }, {
        "_id": ObjectId("5912c71d0520df77f0c2c186"),
        "email_id": "51",
        "unread": "false",
        "__v": NumberLong(0),
        "tag_id": [
            "2", "1"
        ]
    }

预期结果：

{
    "data": [{
        "tag_id": "1",
        "count_email": 1,(count of email on the basis of tag_id)
        "unread": 9(count the unread on the basis of output of tag_id)
    }, {
        "tag_id": "3",
        "count_email": 45,
        "unread": 3
    }, {
        "tag_id": "2",
        "count_email": 5,
        "unread": 4
    }, {
        "id": null,
        "count_email": 52,
        "unread": 35
    }]
}

Answer 1

试试下面的代码 参考-https://docs.mongodb.com/manual/reference/operator/aggregation/eq/

https://docs.mongodb.com/manual/reference/operator/aggregation/cond/

https://docs.mongodb.com/manual/reference/operator/aggregation/group/

DB.aggregate([
                {$project:
                {
                    tag_id: '$tag_id',
                    unreadcount: { $cond: [ { $eq: [ '$unread', 'true' ] }, 1, 0 ] }
                }},
                { $group: {
                    _id: '$tag_id',
                    unread: { $sum: '$unreadcount'},
                }}
            ], function (err, results) {
                console.log(results);
            })