【发布时间】:2020-05-11 04:37:15
【问题描述】:
我正在使用 mongodb 数据库并希望在 2 collections 上应用 $lookup 但有特定条件。
我有一个collection 像这样命名为Company
new Schema({
name: String,
benefit: String,
benefitDesc: String,
company_url: String,
logoUrl: String,
coverUrl: String,
desc: String,
createdAt: String,
categoryId: { type: Schema.Types.ObjectId, ref: 'categories' },
})
另一个collection 像这样命名为Referrallinks
new Schema({
referral_link: String,
referral_code: String,
isLink: Number,
offer_name: String,
offer_desc: String,
user_email: String,
companyId: { type: Schema.Types.ObjectId, ref: 'companies' },
addedByAdmin: { type: Boolean, default: true },
number_of_clicks: Number,
referral_country: String,
link_status: String,
categoryId: { type: Schema.Types.ObjectId, ref: 'categories' },
number_of_clicks: { type: Number, default: 0 },
createdAt: String,
updatedAt: String,
userId: { type: Schema.Types.ObjectId, ref: 'users' }
})
现在当我应用这个$lookup
Company.aggregate([{
$lookup: {
from: 'referrallinks',
localField: '_id',
foreignField: 'companyId',
as: 'referrals'
}
}])
所以我让所有referrals 的公司都像array 这样
[
{name : '', benefit : '', benefiDesc : '', referrals : []},
{name : '', benefit : '', benefiDesc : '', referrals : []},
{name : '', benefit : '', benefiDesc : '', referrals : []},
.....
]
但我希望每个 company's referrals 数组中只有具有 link_status 值 Approved 的推荐。
**注意:**我也想拥有 category 和 company 的对象并尝试了这种聚合,但它给出了错误
{
from: "referrallinks",
let: { company_id: "$_id" },
pipeline: [{
$match: {
$expr: {
$and: [
{ $eq: [ "$$company_id", "$companyId" ] },
{ $eq: [ "$link_status", "Approved" ] }
]
}
},
{
$lookup : {
from : "categories",
let : {"category_id" : "$categoryId"},
pipeline : [
{
$match : {
$expr : {
$eq : ["category_id","$_id"]
}
}
}
],
as : "company"
}
}
}],
as: "referrals"
}
我怎样才能做到这一点?
【问题讨论】:
标签: mongodb mongoose nosql aggregation-framework lookup