par*_*ars 2 mongodb aggregation-framework
假设我有 3 个集合,cars, bikes, vehicles。
cars 集合为:
{
{
"_id": "car1",
"carBrand": "Audi",
"color": "blue"
},
{
"_id": "car2",
"carBrand": "BMW",
"color": "white"
}
}
Run Code Online (Sandbox Code Playgroud)
bikes 集合为:
{
{
"_id": "bike1",
"bikeBrand": "Audi",
"color": "red"
},
{
"_id": "bike2",
"carBrand": "BMW",
"color": "white"
}
}
Run Code Online (Sandbox Code Playgroud)
并且该 vehicles 集合实际上只有对cars和bikes集合的引用
{
{
"_id": "vehicle1",
"vehicleType": "cars",
"vehicleId": "car1"
},
{
"_id": "vehicle2",
"vehicleType": "cars",
"vehicleId": "car2"
},
{
"_id": "vehicle3",
"vehicleType": "bikes",
"vehicleId": "bike1"
},
{
"_id": "vehicle4",
"vehicleType": "bikes",
"vehicleId": "bike2"
},
}
Run Code Online (Sandbox Code Playgroud)
我想加入vehicles收藏cars和bikes收藏。我试图将"$vehicleType"变量设置为$lookup'sfrom字段。然而,它并没有像我预期的那样工作。它根本不连接表。没有错误。
db.collection.aggregate([{
$lookup: {
from: "$vehicleType",
localField: "vehicleId",
foreignField: "_id",
as: "vehicleDetails"
}
}]);
Run Code Online (Sandbox Code Playgroud)
我期待有这样的结果
{
{
"_id": "vehicle1",
"vehicleType": "cars",
"vehicleId": "car1",
"vehicleDetails": {
"_id": "car1",
"carBrand": "Audi",
"color": ""
}
},
{
"_id": "vehicle2",
"vehicleType": "cars",
"vehicleId": "car2",
"vehicleDetails": {
"_id": "car2",
"carBrand": "BMW",
"color": "white"
},
{
"_id": "vehicle3",
"vehicleType": "bikes",
"vehicleId": "bike1",
"vehicleDetails": {
"_id": "bike1",
"bikeBrand": "Audi",
"color": "red"
}
},
{
"_id": "vehicle4",
"vehicleType": "bikes",
"vehicleId": "bike2",
"vehicleDetails": {
"_id": "bike2",
"carBrand": "BMW",
"color": "white"
}
},
}
Run Code Online (Sandbox Code Playgroud)
如果汽车和自行车没有通用 ID,您可以在单独的数组中依次查找,然后将它们与$setUnion 组合:
db.vehicles.aggregate([
{$lookup: {
from: "cars",
localField: "vehicleId",
foreignField: "_id",
as: "carDetails"
}},
{$lookup: {
from: "bikes",
localField: "vehicleId",
foreignField: "_id",
as: "bikeDetails"
}},
{$project: {
vehicleType: 1,
vehicleId: 1,
vehicleDetails:{$setUnion: [ "$carDetails", "$bikeDetails" ]}
}},
{$project: {
carDetails:0,
bikeDetails:0,
}}
]);
Run Code Online (Sandbox Code Playgroud)
否则,您将需要在查找前使用$facet按类型过滤车辆:
db.vehicles.aggregate([
{
$facet: {
"cars": [
{$match: {"vehicleType": "cars"}},
{$lookup: {
from: "cars",
localField: "vehicleId",
foreignField: "_id",
as: "vehicleDetails"
}},
],
"bikes": [
{$match: {"vehicleType": "bikes"}},
{$lookup: {
from: "bikes",
localField: "vehicleId",
foreignField: "_id",
as: "vehicleDetails"
}}
]
}
},
{$project: {all: {$setUnion: ["$cars", "$bikes"]}}},
{$unwind: "$all"},
{$replaceRoot: { newRoot: "$all" }}
])
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
2409 次 |
| 最近记录: |