[英]How to join three (multipe) collections with $lookup in mongodb?
如何在 mongodb 中使用 $lookup 加入三个(多个)集合?
嗨,我希望加入来自三个集合的数据
用户集合:
[
{
_id:0,
name:"abc",
phone:999999999
},
{
_id:1,
name:"xyz",
phone:888888888
},
]
产品集合:
[
{
_id:"p01",
name:"product-name",
price:1200
},
{
_id:"p02",
name:"product-name1",
price:100
}
]
产品评论集合:
[
{
_id:"pr0",
userId:0,
productId:"p01",
star:4
},
{
_id:"pr1",
userId:1,
productId:"p01",
star:3
}
]
mongodb查询:
product.aggregate([
{
$lookup: {
from: "productreviews",
localField: "_id",
foreignField: "productId",
as: "review",
},
},
{
$lookup: {
from: "users",
localField: "review.userId",
foreignField: "_id",
as: "review.userInfo",
},
},
])
我无法获得我需要的输出。 我怎样才能得到以下输出:
{
product: [
{
_id: "p01",
name: "product-name",
price: 1200,
review: [
{
_id: "pr0",
userId: 0,
productId: "p01",
star: 4,
"userInfo": {
name: "abc",
phone: 999999999
}
},
{
_id: "pr1",
userId: 1,
productId: "p01",
star: 3,
"userInfo": {
"name": "xyz",
"phone": 888888888,
}
},
]
},
{
_id: "p02",
name: "product-name1",
price: 100,
},
]
}
任何帮助表示赞赏! 谢谢你...
db.product.aggregate([
{
$lookup: {
from: "review",
localField: "_id",
foreignField: "productId",
as: "review",
},
},
{
$lookup: {
from: "users",
localField: "review.userId", //See here
foreignField: "_id",
as: "review.userInfo",
},
},
])
本地字段名称是第二次查找中的 userId。
编辑:
添加一个放松阶段
db.product.aggregate([
{
$lookup: {
from: "review",
localField: "_id",
foreignField: "productId",
as: "review",
},
},
{
"$unwind": "$review"
},
{
$lookup: {
from: "users",
localField: "review.userId",
foreignField: "_id",
as: "review.userInfo",
},
},
])
要将文档保留在不匹配的位置,请在展开阶段保留空数组,如下所示
{
$unwind: {
path: "$review",
"preserveNullAndEmptyArrays": true
}
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.