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 }
我想显示具有一些条件的用户列表,并在列表顶部显示在线用户和最喜欢的用户列表,在线用户列表显示得分最高和最喜欢的离线用户。规则如下:
- 如果
online
是1
必须按like
的降序排序。 - 如果
online
为0
且score
为1
必须按score< 的降序排序
。 - 如果
online
是0
并且score
是0
必须按like< 的降序排序
。
所以,结果可能是这样的:
{ "_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
}
])
但我有以下当前结果:
{ "_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 },
可以看到,根据第3点,instance { "_id": 8, "name": "H", "online": 0, "like": 13, "score": 0 }
应该在顶部 score
是 0
最佳答案
首先创建附加列调用点
,值为(1 - online)*score
在此之后对数据进行排序:
在线
descpoint
desc (online = 1point
一直为0,online为0point
为score
)喜欢
desc
你可以使用这个查询
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 } }
]
);
关于mongodb - 排序多个条件MongoDB,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50225986/