我的聚合函数遇到问题。我试图从数据库中获取用户最常见的订单,但我只返回名称和计数。我尝试过使用$project
运算符,但我似乎无法让它返回除其中内容之外的任何内容$group
陈述。
这是我当前的聚合函数:
OrderModel.aggregate(
{$unwind: "$products"},
{$match: { customerID: customerID }},
{$group: { _id: "$products.name", count: {$sum:1}}},
{$project: {name: "$_id", _id:0, count:1, active:1}},
{$sort: {"count" : -1}},
{$limit: 25 })
这只会产生如下输出{"count":10, "name": foo"}
而我想返回整个对象;嵌入式文档等等。有什么想法我哪里出错了吗?
编辑-添加示例文档和预期输出
文档:
{
"charge": {},
"captured": true,
"refunds": [
],
"balance_transaction": "txn_104Ics4QFdqlbCVHAdV1G2Hb",
"failure_message": null,
"failure_code": null,
"amount_refunded": 0,
"customer": "cus_4IZMPAIkEdiiW0",
"invoice": null,
"dispute": null,
"statement_description": null,
"receipt_email": null
},
"total": 13.2,
"userToken": "cus_4IZMPAIkEdiiW0",
"customerID": "10152430176375255",
"_id": "53ad927ff0cb43215821c649",
"__v": 0,
"updated": 20140701082928810,
"created": 20140627154919216,
"messageReceived": false,
"ready": true,
"active": false,
"currency": "GBP",
"products": [
{
"name": "Foo",
"active": true,
"types": [
{
"variants": [
{
"name": "Bar",
"isDefault": false,
"price": 13.2
}
]
}
]
}
]
}
预期结果:
[
{
"name": "Foo",
"active": true,
"types": [
{
"variants": [
{
"name": "Bar",
"isDefault": false
}
]
},
{
"variants": [
{
"name": "Something else",
"isDefault": false
}
]
}
],
"quantity": 10
},
{
"name": "Another product",
"active": true,
"types": [
{
"variants": [
{
"name": "Bar",
"isDefault": false
}
]
}
],
"quantity": 7
}
]
Thanks!