Question

我有这样的数据集：

{
    "_id" : ObjectId("5a4c6fb6993a721b3479a27e"),
    "score" : 8.3,
    "page" : "message",
    "lastmodified" : ISODate("2018-01-03T06:49:19.232Z"),
    "createdate" : ISODate("2018-01-03T05:52:54.446Z"),
    "slug" : [ 
        "@APPLE"
    ],
    "__v" : 0
},
    {
        "_id" : ObjectId("5a4c6fb6993a721b3479a27e"),
        "score" : 9.3,
        "page" : "@BANANA",
        "lastmodified" : ISODate("2018-01-03T06:49:19.232Z"),
        "createdate" : ISODate("2018-01-03T05:52:54.446Z"),
        "slug" : [ 
            "@APPLE"
        ],
        "__v" : 0
    }
{
        "_id" : ObjectId("5a4c6fb6993a721b3479a27e"),
        "score" : 5.3,
        "page" : "@BANANA",
        "lastmodified" : ISODate("2018-01-03T06:49:19.232Z"),
        "createdate" : ISODate("2018-01-03T05:52:54.446Z"),
        "slug" : [ 
            "@BANANA"
        ],
        "__v" : 0
    }

现在我想根据我的过滤器计算得分总和：

@APPLE: 8.3+9.3 = 17.6 i.e @APPLE: 17.6,
@BANANA: 9.3+5.3 = 14.6 i.e @BANANA: 14.6

因此，我必须只选择最近1小时的数据而不是选择整个数据库。所以我的查询是这样的

var newTime = new Date();
newTime.setHours( newTime.getHours() - 1 );
db.Test.find({"lastmodified":{$gt: newTime}})

所以通过这个我只能获得最后1小时的价值。现在我很困惑，我怎么能用过滤器来做总结。我还附上了过滤查询，即

db.Test.find({"lastmodified":{$gt: newTime}}, {$or: [{slug: {$in: ['@APPLE']}}, {page: '@APPLE'}]})

但它没有给出任何东西。任何帮助表示赞赏

Answer 1

尝试此聚合查询...

db.tests.aggregate([{
        "$unwind": "$slug"
    },
    {
        "$group": {
            "_id": "$slug",
            "totalScore": {
                "$sum": "$score"
            }
        }
    }
]);

结果：

{ 
    "_id" : "@BANANA", 
    "totalScore" : 5.3
}
{ 
    "_id" : "@APPLE", 
    "totalScore" : 17.6
}

合并数组和组以为每个组合数组值生成计数

1 个答案: