这是我的收藏:
[
{userId: "u1", data: { score1: 1, score2: 2, score3: 3 }, day: 1},
{userId: "u1", data: { score1: 1, score2: 0, score3: 0 }, day: 2},
{userId: "u1", data: { score1: 5, score2: 3, score3: 2 }, day: 3},
{userId: "u2", data: { score1: 2, score2: 5, score3: 1 }, day: 1},
{userId: "u2", data: { score1: 1, score2: 1, score3: 6 }, day: 2},
{userId: "u2", data: { score1: 3, score2: 5, score3: 3 }, day: 3},
{userId: "u3", data: { score1: 4, score2: 1, score3: 1 }, day: 1},
{userId: "u3", data: { score1: 0, score2: 1, score3: 1 }, day: 2},
{userId: "u3", data: { score1: 0, score2: 1, score3: 10 }, day: 3}
]
我想建立以下排行榜表:
{
score1: [
{"u1": 7}, // sum of all score1 for u1
{"u2": 6}, // sum of all score1 for u2
{"u3": 4}, // sum of all score1 for u3
],
score2: [
{"u2": 11}, // sum of all score2 for u2
{"u1": 5}, // sum of all score2 for u1
{"u3": 3}, // sum of all score2 for u3
],
score3: [
{"u3": 12}, // sum of all score3 for u3
{"u2": 10}, // sum of all score3 for u2
{"u1": 5}, // sum of all score3 for u1
],
}
到目前为止,我可以按userId分组,并为其中的3个计算每个分数的总和:
db.myCollection.aggregate([
{
$group: {
_id: "$userId",
score1: { $sum: "$score1" },
score2: { $sum: "$score2" },
score3: { $sum: "$score3" }
}
}
])
哪个给我:
[
{
_id: "u1",
score1: 7,
score2: 5,
score3: 5
},
{
_id: "u2",
score1: 6,
score2: 11,
score3: 10
},
{
_id: "u3",
score1: 4,
score2: 3,
score3: 12
},
]
如何提取每种分数并建立相应的排行榜?
提前感谢。
$objectToArray
字段上使用data
,然后在$unwind
上使用,因此每个文档有1个用户和1个分数。然后按userId
和data.k
(将包含“ score1”,“ score2”等)分组并计算总和。然后按乐谱名称重新分组,并将带有k:userId, v:<score>
的对象压入数组。然后再次在null
上分组,然后将k:scoreName, v:<object with user scores>
推入阵列。最后,$arrayToObject
将该数组转换为所需的对象:db.collection.aggregate([
{$addFields: {data: {$objectToArray: "$data"}}},
{$unwind: "$data"},
{$group: {
_id: {userId: "$userId", scoreName: "$data.k"},
score: {$sum:"$data.v"}
}},
{$group: {
_id:"$_id.scoreName",
data:{$push:{k:"$_id.userId", v:"$score"}}
}},
{$group: {
_id: null,
scores:{$push:{k:"$_id", v:{$arrayToObject:"$data"}}}
}},
{$replaceRoot:{newRoot:{$arrayToObject:"$scores"}}}
])
Playground