Question

我使用过MongoDB聚合，当我尝试使用sort和limit时，我会得到重复的记录。

如下所示，我有3个文档，其中document1和document2具有相同的标题，document3具有不同的视频标题。

我想按标题对文档进行分组，并按序列集按顺序对分组文档进行排序。

document1和document2将在同一个群组中，并且会获得唯一的文档2，然后我想要将此document2和document3与video_view_count进行排序

/* document1 */
{
    "_id" : ObjectId("580afd565706467c1bbabd70"),
    "serial_episode" : "5",
    "video_view_count" : 50.0,
    "video_data" : [ 
        {
            "video_categories" : [ 
                "Sport"
            ],
            "video_title" : "Zwyci??zca",
            "language_id" : "578f1ec6e494f9400b21fec4"
        }, 
        {
            "video_featured_text" : "",
            "video_categories" : [ 
                "Sport"
            ],
            "video_title" : "Zwyci??zca",
            "language_id" : "578f1ec6e494f9400b21fec3"
        }
    ]
}

/* document2 */
{
    "_id" : ObjectId("580afd565706467c1bbabd71"),
    "serial_episode" : "6",
    "video_view_count" : 10.0,
    "video_data" : [ 
        {
            "video_categories" : [ 
                "Sport"
            ],
            "video_title" : "Zwyci??zca",
            "language_id" : "578f1ec6e494f9400b21fec4"
        }, 
        {
            "video_featured_text" : "",
            "video_categories" : [ 
                "Sport"
            ],
            "video_title" : "Zwyci??zca",
            "language_id" : "578f1ec6e494f9400b21fec3"
        }
    ]
}

/* document3 */
{
    "_id" : ObjectId("580afd565706467c1bbabd72"),
    "serial_episode" : "",
    "video_view_count" : 11.0,
    "video_data" : [ 
        {
            "video_categories" : [ 
                "Sport"
            ],
            "video_title" : "Zwyci??zca123",
            "language_id" : "578f1ec6e494f9400b21fec4"
        }, 
        {
            "video_featured_text" : "",
            "video_categories" : [ 
                "Sport"
            ],
            "video_title" : "Zwyci??zca123",
            "language_id" : "578f1ec6e494f9400b21fec3"
        }
    ]
}

Expexcted Output：

我希望结果为document3，然后是document2，因为docuemnt1-2组合在一起以生成uniqe document2，docuemnt2包含最新一集。我需要将document2和document3与视频观看次数进行比较：

/* document3 */
{
    "_id": ObjectId("580afd565706467c1bbabd72"),
    "serial_episode": "",
    "video_view_count": 11,
    "video_data": [
        {
            "video_categories": [
                "Sport"
            ],
            "video_title": "Zwyci??zca123",
            "language_id": "578f1ec6e494f9400b21fec4"
        },
        {
            "video_featured_text": "",
            "video_categories": [
                "Sport"
            ],
            "video_title": "Zwyci??zca123",
            "language_id": "578f1ec6e494f9400b21fec3"
        }
    ]
},
/* document3 */
{
    "_id": ObjectId("580afd565706467c1bbabd71"),
    "serial_episode": "6",
    "video_view_count": 10,
    "video_data": [
        {
            "video_categories": [
                "Sport"
            ],
            "video_title": "Zwyci??zca",
            "language_id": "578f1ec6e494f9400b21fec4"
        },
        {
            "video_featured_text": "",
            "video_categories": [
                "Sport"
            ],
            "video_title": "Zwyci??zca",
            "language_id": "578f1ec6e494f9400b21fec3"
        }
    ]
}

当前聚合操作：

// Grouping that I have implemented 
var group = {   
    "$group": { 
        "_id":" $video_data.video_title",
        "video_rating" : { $first:" $video_rating" },
        "serial_episode" : { $first: "$serial_episode" },
        "video_view_count": { $first: "$video_view_count" },
    }
};

// Aggregate function to get that videos    
videos.aggregate([ 
    { $match: { "video_data.video_categories": query.category_name } },
    { $unwind: "$video_data" },
    { $sort: { video_view_count: -1 } },
    { $sort:{ serial_episode: -1 } },
    group,
    { $sort:{ video_view_count: -1 } },
    { $skip: skipData },
    { $limit: 10 }
], function(error, output){});

Answer 1

你为什么要解开它？移除展开，它会将文档与数组长度相乘..

使用mongodb聚合进行数据复制

1 个答案: