这是我的收藏:
{
"_id" : 10926400,
"votes": 131,
"author": "Jesse",
"comments" : [
{
"id" : 1,
"votes": 31,
"author": "Mirek"
},
{
"id": 2,
"votes": 13,
"author": "Leszke"
}
]
},
{
"_id" : 10926401,
"votes": 75,
"author": "Mirek",
"comments" : [
{
"id" : 1,
"votes": 17,
"author": "Jesse"
},
{
"id": 2,
"votes": 29,
"author": "Mirek"
}
]
}
我希望每个$sum
的{{1}}和votes
的值comments.votes
预期输出(author
):
sort $votes: -1
答案 0 :(得分:4)
不是立即可见但可能。您需要做的是将顶级文档与注释数组合并而不重复。这是首先将内容作为两个数组加入单个数组,然后$unwind
对内容进行分组的方法:
db.collection.aggregate([
{ "$group": {
"_id": "$_id",
"author": {
"$addToSet": {
"id": "$_id",
"author": "$author",
"votes": "$votes"
}
},
"comments": { "$first": "$comments" }
}},
{ "$project": {
"combined": { "$setUnion": [ "$author", "$comments" ] }
}},
{ "$unwind": "$combined" },
{ "$group": {
"_id": "$combined.author",
"votes": { "$sum": "$combined.votes" }
}},
{ "$sort": { "votes": -1 } }
])
给出了输出:
{ "_id" : "Jesse", "votes" : 148 }
{ "_id" : "Mirek", "votes" : 135 }
{ "_id" : "Leszke", "votes" : 13 }
即使跳过第一个$group
阶段并以不同的方式组合数组:
db.collection.aggregate([
{ "$project": {
"combined": {
"$setUnion": [
{ "$map": {
"input": { "$literal": ["A"] },
"as": "el",
"in": {
"author": "$author",
"votes": "$votes"
}
}},
"$comments"
]
}
}},
{ "$unwind": "$combined" },
{ "$group": {
"_id": "$combined.author",
"votes": { "$sum": "$combined.votes" }
}},
{ "$sort": { "votes": -1 } }
])
那些使用MongoDB 2.6中引入的$setUnion
甚至$map
等运算符。这使得它更简单,但它仍然可以在缺少这些运算符的早期版本中完成,遵循相同的原则:
db.collection.aggregate([
{ "$project": {
"author": 1,
"votes": 1,
"comments": 1,
"type": { "$const": ["A","B"] }
}},
{ "$unwind": "$type" },
{ "$unwind": "$comments" },
{ "$group": {
"_id": {
"$cond": [
{ "$eq": [ "$type", "A" ] },
{
"id": "$_id",
"author": "$author",
"votes": "$votes"
},
"$comments"
]
}
}},
{ "$group": {
"_id": "$_id.author",
"votes": { "$sum": "$_id.votes" }
}},
{ "$sort": { "votes": -1 } }
])
$const
未记录,但存在于聚合框架所在的所有MongoDB版本中(来自2.2)。 MongoDB 2.6引入了$literal
,它基本上链接到相同的底层代码。这里有两种情况用于为数组提供模板元素,或者引入一个数组来展开,以便在两个动作之间提供“二元选择”。
答案 1 :(得分:2)
您可以按以下方式汇总结果:
Unwind
条数组。Group
记录在一起首先计算投票总和
每位作者在评论中收到。同时保留原件
帖子很明智。Unwind
。project
每位作者的总和。Sort
通过作者的姓名和投票。代码:
db.collection.aggregate([
{$unwind:"$comments"},
{$group:{"_id":null,
"comments":{$push:"$comments"},
"post":{$addToSet:{"author":"$author",
"votes":"$votes"}}}},
{$unwind:"$comments"},
{$group:{"_id":"$comments.author",
"votes":{$sum:"$comments.votes"},
"post":{$first:"$post"}}},
{$unwind:"$post"},
{$project:{"_id":1,
"votes":{$cond:[{$eq:["$_id","$post.author"]},
{$add:["$votes","$post.votes"]},
"$votes"]}}},
{$sort:{"_id":-1,"votes":-1}},
{$group:{"_id":"$_id","votes":{$first:"$votes"}}}
])
样本o / p:
{ "_id" : "Leszke", "votes" : 13 }
{ "_id" : "Jesse", "votes" : 148 }
{ "_id" : "Mirek", "votes" : 135 }