mongodb - 如何在 MongoDB 聚合管道中执行嵌套 "joins"(加入 3 个或更多集合)？

Question

假设我们在 MongoDB 中有 3 个假设集合:customers、orders 和 orderItems。

每个客户有多个订单，每个订单有多个订单项。

这是这 3 个集合的一些示例数据:

客户

[
    {
        customer_id: 1,
        name: "Jim Smith",
        email: "jim.smith@example.com"
    },
    {
        customer_id: 2,
        name: "Bob Jones",
        email: "bob.jones@example.com"
    }
]

订单

[
    {
        order_id: 1,
        customer_id: 1
    },
    {
        order_id: 2,
        customer_id: 1
    }
]

订单商品

[
    {
        order_item_id: 1,
        name: "Foo",
        price: 4.99,
        order_id: 1
    },
    {
        order_item_id: 2,
        name: "Bar",
        price: 17.99,
        order_id: 1
    },
    {
        order_item_id: 3,
        name: "baz",
        price: 24.99,
        order_id: 2
    }
]

期望的结果

我如何编写我的聚合管道，以便返回的结果看起来像这样？

[
    {
        customer_id: 1,
        name: "Jim Smith",
        email: "jim.smith@example.com"
        orders: [
            {
                order_id: 1,
                items: [
                    {
                        name: "Foo",
                        price: 4.99
                    },
                    {
                        name: "Bar",
                        price: 17.99
                    }
                ]
            },
            {
                order_id: 2,
                items: [
                    {
                        name: "baz",
                        price: 24.99
                    }
                ]
            }
        ]
    },
    {
        customer_id: 2,
        name: "Bob Jones",
        email: "bob.jones@example.com"
        orders: []
    }
]

Answer 1

使用 lookup with pipeline 进行嵌套查找,

• $lookup 和 orders 集合，
  • let，定义来自主集合的变量 customer_id，使用 $$ 像 $ 访问管道内的这个引用变量$customer_id,
  • pipeline 可以像我们在根级管道中一样添加管道阶段
  • $expr 每当我们匹配内部字段时，它需要表达式匹配条件，所以 $$customer_id 是在 let 中声明的父集合字段，并且$customer_id 是子集合/当前集合的字段
• $lookup 与 orderitems 集合

db.customers.aggregate([
  {
    $lookup: {
      from: "orders",
      let: { customer_id: "$customer_id" },
      pipeline: [
        { $match: { $expr: { $eq: ["$$customer_id", "$customer_id"] } } },
        {
          $lookup: {
            from: "orderitems",
            localField: "order_id",
            foreignField: "order_id",
            as: "items"
          }
        }
      ],
      as: "orders"
    }
  }
])

Playground

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Tip:

Several joins considered as bad practice in NoSQL, I would suggest if you could add your order items in orders collection as array, you can save one join process for orderitems, see improved version in playground