Question

我在mongodb：中有集合，存储为：

{"tag":"count1","value":100,"ts":1544423706} {"tag":"count2","value":1002,"ts":1544423706} {"tag":"count1","value":101,"ts":1544423806} {"tag":"count2","value":1003,"ts":1544423806} {"tag":"count1","value":102,"ts":1544423906} {"tag":"count2","value":1004,"ts":1544423906}

所以我的问题是如何从“标签”中获取结果为count1，“ ts”大于1544423800的第一项。如我所愿：我想找到的结果是：

{"tag":"count1","value":101,"ts":1544423806} {"tag":"count2","value":1003,"ts":1544423806}

我是否需要使用聚合将标签分组，然后获得比给定的“ ts”大的第一项，所以我是MongoDB中聚合功能的新手。

  db.index.aggregate([{"$match": {"tag": {"$in":["count1","count2"]},"ts": {"$gt":1544423800}}
                        },
            {"$group": {"_id": "$tag",
                        "tags": {"$push": "$$ROOT"}}
                        },
        ])

结果不是每个标签都有一项，所以我想限制一项，我该怎么做

谢谢，我已通过以下方式解决了此问题：

db.index.aggregate(

[
            {"$match": {"tag": {"$in":["count1","count2"]},
                                             "ts": {"$gt":1545730000}}
                        },
            {"$group": {"_id": "$tag",
                        "value": {"$first": "$value"}
                                             }
                        },

        ]

)

mongo聚合组并找到一个

0 个答案: