MongoDB在不进行Unwinding的情况下聚合组和计数嵌套数组

Question

是否可以创建一个aggerate，对嵌套的对象数组进行分组和计数，而无需展开它。

EG，拿这样的东西：

"things": [
{
"something": "else",
"stuff": "I don't want",
"a_thing": [
{
"thing": "a record",
"cat": "type A"
},
{
"thing": "an other record",
"cat": "type A"
},
{
"thing": "yet other record",
"cat": "type B"
},
{
"thing": "more records",
"cat": "type A"
},
{
"thing": "last record",
"cat": "type C"
}
]
}
]

结果是这样的：

"things": [{
{ "Grand_Total": 5 },
[{
"type A":
{ "sub_total": 3 },
[{
{ "thing": "a record" },
{ "thing": "an other record" },
{ "thing": "more records" }
}],
"type B":
{ "sub_total": 1 },
[{
{ "thing": "yet other record" }
}],
"type C":
{ "sub_total": 1 },
[{
{ "thing": "last record" }
}]
}]
}]

我一直在使用$project和$filter来减少一定条件下的记录，但我不确定如果没有放松阶段，分组和计数是否可行。

谢谢。

Answer 1

一个选项是：

1. 创建一个键数组
2. 收集每个密钥的a_thing项
3. 格式化每个键的数据以匹配$arrayToObject
4. 使用$arrayToObject格式化答案

db.collection.aggregate([
{$set: {keys: {$setUnion: ["$a_thing.cat"]}}},
{$project: {
res: {$map: {
input: "$keys",
as: "key",
in: {
k: "$$key",
v: {$reduce: {
input: "$a_thing",
initialValue: [],
in: {$concatArrays: [
"$$value",
{$cond: [
{$eq: ["$$this.cat", "$$key"]},
[{thing: "$$this.thing"}],
[]
]}
]}
}}
}
}}
}},
{$project: {
res: {$map: {
input: "$res",
in: {k: "$$this.k", v: {sub_total: {$size: "$$this.v"}, values: "$$this.v"}}
}}
}},
{$project: {res: {"$arrayToObject": "$res"}}}
])

看看它是如何在操场上工作的例子