聚合文档多级
聚合文档多级
提问于 2018-06-05 21:32:20
现在考虑一下这个例子，我有一个文档，包含下面的类似集合的结构。下面是订单集合
{
"_id" : ObjectId("5788fcd1d8159c2366dd5d93"),
"color" : "Blue",
"code" : "1",
"category_id" : ObjectId("5693d170a2191f9020b8c815"),
"description" : "julia tried",
"name" : "Order1",
"brand_id" : ObjectId("5b0e52f058b8287a446f9f05")
}还有一个Brand和Category集合。这是类别集合
{
"_id" : ObjectId("5693d170a2191f9020b8c815"),
"name" : "Category1",
"created_at" : ISODate("2016-01-11T20:32:17.832+0000"),
"updated_at" : ISODate("2016-01-11T20:32:17.832+0000"),
}品牌收藏
{
"_id" : ObjectId("5b0e52f058b8287a446f9f05"),
"name" : "brand1",
"description" : "brand1",
"updated_at" : ISODate("2017-07-05T09:18:13.951+0000"),
"created_at" : ISODate("2017-07-05T09:18:13.951+0000"),
}在应用聚合之后，现在应该会产生以下格式：
{
'brands': [
{
_id: '*******'
name: 'brand1',
categories: [
{
_id: '*****',
name: 'category_name1',
orders: [
{
_id: '*****',
title: 'order1'
}
]
}
]
}
]
}回答 1
Stack Overflow用户
回答已采纳
发布于 2018-06-06 00:40:07
你可以尝试下面的聚合：
db.brand.aggregate([
{
$lookup: {
from: "order",
localField: "_id",
foreignField: "brand_id",
as: "orders"
}
},
{
$unwind: "$orders"
},
{
$lookup: {
from: "category",
localField: "orders.category_id",
foreignField: "_id",
as: "categories"
}
},
{
$unwind: "$categories"
},
{
$group: {
_id: "$_id",
name: { $first: "$name" },
description: { $first: "$description" },
updated_at: { $first: "$updated_at" },
created_at: { $first: "$created_at" },
categories: { $addToSet: "$categories" },
orders: { $addToSet: "$orders" }
}
},
{
$addFields: {
categories: {
$map: {
input: "$categories",
as: "category",
in: {
$mergeObjects: [
"$$category", {
orders: [ {
$filter: {
input: "$orders",
as: "order",
cond: { $eq: [ "$$category._id", "$$order.category_id" ] }
}
} ]
} ]
}
}
}
}
},
{
$project: {
orders: 0
}
}
])基本上，您必须使用$lookup两次才能根据brand_id和category_id字段“合并”所有这些集合中的数据。由于您期望在brands的categories中使用orders，因此可以对这两个数组使用$unwind，然后使用$group获得以下形状：
{
"_id" : ObjectId("5b0e52f058b8287a446f9f05"),
"name" : "brand1",
"description" : "brand1",
"updated_at" : ISODate("2017-07-05T09:18:13.951Z"),
"created_at" : ISODate("2017-07-05T09:18:13.951Z"),
"categories" : [
{
"_id" : ObjectId("5693d170a2191f9020b8c814"),
"name" : "Category1",
"created_at" : ISODate("2016-01-11T20:32:17.832Z"),
"updated_at" : ISODate("2016-01-11T20:32:17.832Z")
}
],
"orders" : [
{
"_id" : ObjectId("5788fcd1d8159c2366dd5d93"),
"color" : "Blue",
"code" : "1",
"category_id" : ObjectId("5693d170a2191f9020b8c814"),
"description" : "julia tried",
"name" : "Order1",
"brand_id" : ObjectId("5b0e52f058b8287a446f9f05")
}
]
}现在，您有了brand1及其所有子类别，以及应该放在其中一个类别中的所有订单。唯一的问题是如何在categories中“嵌套”orders。要做到这一点，一种方法可能是$map，您可以将每个类别与匹配该类别的所有订单合并(使用$mergeObjects，您不必指定categories对象中的所有属性)。
要将category与orders匹配，您可以在orders阵列上执行$filter。
然后，您可以删除orders，因为它们嵌套在类别中，因此您不再需要它们。
编辑: 3.4版本
在MongoDB 3.4中，你不能使用$mergeObjects，所以你应该指定``categories的所有属性：
db.brand.aggregate([
{
$lookup: {
from: "order",
localField: "_id",
foreignField: "brand_id",
as: "orders"
}
},
{
$unwind: "$orders"
},
{
$lookup: {
from: "category",
localField: "orders.category_id",
foreignField: "_id",
as: "categories"
}
},
{
$unwind: "$categories"
},
{
$group: {
_id: "$_id",
name: { $first: "$name" },
description: { $first: "$description" },
updated_at: { $first: "$updated_at" },
created_at: { $first: "$created_at" },
categories: { $addToSet: "$categories" },
orders: { $addToSet: "$orders" }
}
},
{
$addFields: {
categories: {
$map: {
input: "$categories",
as: "category",
in: {
_id: "$$category._id",
name: "$$category.name",
created_at: "$$category.created_at",
updated_at: "$$category.updated_at",
orders: [
{
$filter: {
input: "$orders",
as: "order",
cond: { $eq: [ "$$category._id", "$$order.category_id" ] }
}
}
]
}
}
}
}
},
{
$project: {
orders: 0
}
}
])页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50701420
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