MongoDB 集合也不是实际的模式,我假设:
如果是这种情况,您需要的是一个 MongoDB Aggregate,您可以在“sells”集合上运行,“join”与“product”集合并按类别 ID 分组。
这样做的基本聚合将遵循以下几行:
sells.aggregate([
{
$lookup: {
from: "product",
localField: "productId",
foreignField: "id",
as: "ProductData"
}
},
{
$unwind: "$ProductData"
},
{
$group: {
_id: "$ProductData.catId",
total: { $sum: "$value" }
}
}
]);
如果您还想在聚合后获取类别名称,您只需在与类别集合连接的管道末尾插入另一个 $lookup:
sells.aggregate([
{
$lookup: {
from: "product",
localField: "productId",
foreignField: "id",
as: "ProductData"
}
},
{
$unwind: "$ProductData"
},
{
$group: {
_id: "$ProductData.catId",
total: { $sum: "$value" }
}
},
{
$lookup: {
from: "category",
localField: "_id",
foreignField: "id",
as: "CategoryData"
}
},
{
$unwind: "$CategoryData"
},
{
$project: {
name: "$CategoryData.name",
total: 1
}
}
]);
编辑(在评论中添加新案例请求):
db.getCollection('product').aggregate([
{
$lookup: {
from: "sells",
localField: "id",
foreignField: "productId",
as: "SalesData"
}
},
{
$unwind:
{
path: "$SalesData",
preserveNullAndEmptyArrays: true
}
},
{
$project: {
catId: 1,
value: "$SalesData.value"
}
},
{
$group: {
_id: "$catId",
total: { $sum: "$value" }
}
},
{
$lookup: {
from: "category",
localField: "_id",
foreignField: "id",
as: "CategoryData"
}
},
{
$unwind: "$CategoryData"
},
{
$project: {
name: "$CategoryData.name",
total: 1
}
}
]);