我想创建一个新文档，该文档从所有具有相同buyer的订单中获取所有cartItem。如果它没有一对（就像伦纳德一样）它会创建新的文档，但状态为"orderId" : "merged"。
例如： 这是需要的情况，当一些客户将做出几个不同的订单，但我只需要给出一个合并的配方。
收藏orders：
输入
{
"_id" : "001",
"buyer": "Sheldon"
"cartItems" : [
{
"itemName" : "Water",
"itemPrice" : 3
}
],
"totalCost" : 3
},
{
"_id" : "002",
"buyer" : "Sheldon",
"cartItems" : [
{
"itemName" : "Milk",
"itemPrice" : 2
}
],
"totalCost" : 2
},
{
"_id" : "003",
"buyer" : "Sheldon",
"cartItems" : [
{
"itemName" : "Butter",
"itemPrice" : 4
}
],
"totalCost" : 4
},
{
"_id" : "004",
"buyer" : "Leonard",
"cartItems" : [
{
"itemName" : "Water",
"itemPrice" : 3
}
],
"totalCost" : 3
}
的输出
{
"_id" : "003_new",
"buyer" : "Sheldon",
"cartItems" : [
{
"itemName" : "Water",
"itemPrice" : 3
},
{
"itemName" : "Milk",
"itemPrice" : 2
},
{
"itemName" : "Butter",
"itemPrice" : 4
}
],
"totalCost" : 9,
"orderId" : "merged"
},
{
"_id" : "004_new",
"buyer" : "Leonard",
"cartItems" : [
{
"itemName" : "Water",
"itemPrice" : 3
}
],
"totalCost" : 3,
"orderId" : "merged"
}
如果你在JS中提供示例，会更好。
答案 0 :(得分：1)
db.orders.aggregate([
{$sort: {_id: 1, buyer: 1}},
{$unwind: '$cartItems'},
{$group: {_id: '$buyer', cartItems: {$push: '$cartItems'},
totalCost: {$sum: '$totalCost'},
id: {$last: {$concat: ["$_id", "_", "new" ]}},
buyer: {$last: '$buyer'}}},
{$addFields: {orderId: 'merged', _id: '$id'}},
{$project: {"id": 0 }}])