Question

我有用户的集合，这是以下文件：

{ "_id": 1, "name": "A", "online": 1, "like": 10, "score": 1 },
{ "_id": 2, "name": "B", "online": 0, "like": 9, "score": 0 },
{ "_id": 3, "name": "C", "online": 0, "like": 8, "score": 1 },
{ "_id": 4, "name": "D", "online": 1, "like": 8, "score": 0 },
{ "_id": 5, "name": "E", "online": 1, "like": 7, "score": 1 },
{ "_id": 6, "name": "F", "online": 0, "like": 10, "score": 1 },
{ "_id": 7, "name": "G", "online": 0, "like": 5, "score": 0 },
{ "_id": 8, "name": "H", "online": 0, "like": 13, "score": 0 }
{ "_id": 9, "name": "I", "online": 0, "like": 6, "score": 0 }

我想显示一些标准的用户列表，并通过一些条件，在线用户和最受欢迎的用户列表排序，在线用户列表显示离线用户得分最多＆amp;最喜欢的。以下规则：

如果online 1必须按like的降序排序。
如果online为0且score为1，则必须按score的降序进行排序。
如果online为0且score为0，则必须按like的降序进行排序。

所以，结果可能是：

{ "_id": 1, "name": "A", "online": 1, "like": 10, "score": 1 },
{ "_id": 4, "name": "D", "online": 1, "like": 8, "score": 0 },
{ "_id": 5, "name": "E", "online": 1, "like": 7, "score": 1 },
{ "_id": 6, "name": "F", "online": 0, "like": 10, "score": 1 },
{ "_id": 3, "name": "C", "online": 0, "like": 8, "score": 1 },
{ "_id": 8, "name": "H", "online": 0, "like": 13, "score": 0 }
{ "_id": 2, "name": "B", "online": 0, "like": 9, "score": 0 },
{ "_id": 9, "name": "I", "online": 0, "like": 6, "score": 0 },
{ "_id": 7, "name": "G", "online": 0, "like": 5, "score": 0 }

我已经完成了第2点，我的查询如下：

db.users.aggregate([
{
   $project :
       {
           "id" : 1,
           "name" : 1,
           "online: 1,
           "like" : 1,
           "score" : 1,
           "sort" : {
               $cond:
                   {
                       "if" :
                           {
                               $eq : ["$online", true]
                           },
                       "then" : "$like",
                       "else" : "$score"
                   }
           }
       }
},
{
   $sort :
       {
           "online" : -1,
           "sort" : -1,
           "id" : 1
       }
},
{
   $skip : 0
},
{
   $limit : 9
}
])

但我现在有以下结果：

{ "_id": 1, "name": "A", "online": 1, "like": 10, "score": 1 },
{ "_id": 4, "name": "D", "online": 1, "like": 8, "score": 0 },
{ "_id": 5, "name": "E", "online": 1, "like": 7, "score": 1 },
{ "_id": 6, "name": "F", "online": 0, "like": 10, "score": 1 },
{ "_id": 3, "name": "C", "online": 0, "like": 8, "score": 1 },
{ "_id": 2, "name": "B", "online": 0, "like": 9, "score": 0 },
{ "_id": 7, "name": "G", "online": 0, "like": 5, "score": 0 },
{ "_id": 8, "name": "H", "online": 0, "like": 13, "score": 0 }
{ "_id": 9, "name": "I", "online": 0, "like": 6, "score": 0 },

您可以看到，基于第3点，实例{ "_id": 8, "name": "H", "online": 0, "like": 13, "score": 0 }应位于首位，score为0

Answer 1

首先创建值point

的其他列调用(1 - online)*score

经过以下排序后的数据：

online desc
point desc（在线= 1 point始终为0，在线为0 point为score）
like desc

您可以使用此查询

    db.yourtable.aggregate(
            [ 
                { $project:{
                    "id" : 1,
                    "name" : 1,
                    "online": 1,
                    "like" : 1,
                    "score" : 1,
                    point: { $multiply: [ 
                                    {$subtract: [1,"$online"]}                      
                                    , "$score"
                            ]}

                    }
                }
                ,{ $sort : { online: -1, point : -1, like : -1 } }
            ]
        );

Answer 2

请查看以下查询：

{{1}}

排序多个条件MongoDB

2 个答案: