MongoDB 中 3 个级别的深度链接

Question

我需要从数据库中检索整个单个对象层次结构作为 JSON。我决定使用带有 $lookup 支持的 MongoDB。

所以我有三个集合：

1.餐厅

{
    "_id" : ObjectId("57c6a02021c8b8d04443aa7c"),
    "name" : "KFC",
    "city" : "Lahore",
    "area" : "Johar town",
    "min_order" : "500",
    "del_time" : "1 hour",
    "del_fees" : "5",
    "pre_order" : "no",
    "rating_star" : "0",
    "rating_no" : 0,
    "status" : "1",
    "working_hours" : {
        "monday" : "2pm 6am",
        "tuesday" : "6am 8pm",
        "wednesday" : "9pm 5pm",
        "thursday" : "3pm 6pm",
        "friday" : "1pm 9pm",
        "saturday" : "4pm 5pm",
        "sunday" : "4pm 8pm"
    },
    "cuisines" : [
        "fast food, pizza"
    ],
    "payments" : [
        "cash,credit card,paypal"
    ]
}

2。类别

"categories" : [
    {
        "_id" : ObjectId("57c6a05421c8b8d04443aa7e"),
        "restaurant_id" : ObjectId("57c6a02021c8b8d04443aa7c"),
        "name" : "Breakfast",
        "category_id" : "1",
        "__v" : 0
    },
    {
        "_id" : ObjectId("57c6a06021c8b8d04443aa7f"),
        "restaurant_id" : ObjectId("57c6a02021c8b8d04443aa7c"),
        "name" : "Special",
        "category_id" : "2",
        "__v" : 0
    }
]

3。项目

"items" : [
            {
                "_id" : ObjectId("57c6da8cc8d053e8310dfe65"),
                "restaurant_id" : ObjectId("57c6a02021c8b8d04443aa7c"),
                "category_id" : ObjectId("57c6a05421c8b8d04443aa7e"),
                "name" : "Water",
                "price" : "55",
                "rating" : "0",
                "__v" : 0
            },
            {
                "_id" : ObjectId("57c6da96c8d053e8310dfe66"),
                "restaurant_id" : ObjectId("57c6a02021c8b8d04443aa7c"),
                "category_id" : ObjectId("57c6a05421c8b8d04443aa7e"),
                "name" : "Milk",
                "price" : "55",
                "rating" : "0",
                "__v" : 0
            }
        ]

我想要这样的 Json 响应输出

{
    "_id" : ObjectId("57c6a02021c8b8d04443aa7c"),
    "name" : "KFC",
    "city" : "Lahore",
    "area" : "Johar town",
    "min_order" : "500",
    "del_time" : "1 hour",
    "del_fees" : "5",
    "pre_order" : "no",
    "rating_star" : "0",
    "rating_no" : 0,
    "status" : "1",
    "working_hours" : {
        "monday" : "2pm 6am",
        "tuesday" : "6am 8pm",
        "wednesday" : "9pm 5pm",
        "thursday" : "3pm 6pm",
        "friday" : "1pm 9pm",
        "saturday" : "4pm 5pm",
        "sunday" : "4pm 8pm"
    },
    "cuisines" : [
        "fast food, pizza"
    ],
    "payments" : [
        "cash,credit card,paypal"
    ],
    "__v" : 0,
    "categories" : [{
        "_id" : ObjectId("57c6a05421c8b8d04443aa7e"),
        "restaurant_id" : ObjectId("57c6a02021c8b8d04443aa7c"),
        "name" : "Breakfast",
        "category_id" : "1",
        "__v" : 0,
        "items" : [
            {
                "_id" : ObjectId("57c6da8cc8d053e8310dfe65"),
                "restaurant_id" : ObjectId("57c6a02021c8b8d04443aa7c"),
                "category_id" : ObjectId("57c6a05421c8b8d04443aa7e"),
                "name" : "Water",
                "price" : "55",
                "rating" : "0",
                "__v" : 0
            },
            {
                "_id" : ObjectId("57c6da96c8d053e8310dfe66"),
                "restaurant_id" : ObjectId("57c6a02021c8b8d04443aa7c"),
                "category_id" : ObjectId("57c6a05421c8b8d04443aa7e"),
                "name" : "Milk",
                "price" : "55",
                "rating" : "0",
                "__v" : 0
            }
        ]
    }]
}

我还是这样的

db.restaurants.aggregate([{
    $lookup: {
        from: "categories",
        localField: "_id",
        foreignField: "restaurant_id",
        as: "categories"
    }
}])

Restaurants 和 categories 工作正常，但我如何处理 categories 中的每个类别项目？

Answer 1

您可以尝试以下聚合管道，记住您需要对创建的查找数组执行 $unwind 操作，以便下一个 $lookup 要应用的操作：

db.restaurants.aggregate([
    {
        "$lookup": {
            "from": "categories",
            "localField": "_id",
            "foreignField": "restaurant_id",
            "as": "categories"
        }
    },
    { "$unwind": "$categories" },
    {
        "$lookup": {
            "from": "items",
            "localField": "categories._id",
            "foreignField": "category_id",
            "as": "categories.items"
        }
    }
 ])

Answer 2

我不认为这是直接可能的。根据documentation on $lookup：

如果你的 localField 是一个数组并且你想匹配元素 在它里面反对一个单一元素的foreignField，你会 需要 $unwind 数组作为聚合管道的一个阶段。

所以，基本上您需要 $unwind 您的类别数组才能查找项目。但是，只要将它重新拼凑起来，您就可以发挥创造力。

例如，这接近你想要的：

db.restaurants.aggregate([
{
    $lookup: {
        from: "categories",
        localField: "_id",
        foreignField: "restaurant_id",
        as: "categories"
    }
},
{
    $unwind: "$categories"
},
{
    $lookup: {
        from: "items",
        localField: "categories._id",
        foreignField: "category_id",
        as: "items"
    }
},
{
    $project:{
        category: {
            elem: "$categories",
            items: "$items"
        }
    }
},
{
    $group: {
        _id: "$_id",
        categories: {$addToSet: "$category"}
    }
},
{
    $lookup: {
        from: "restaurants",
        localField: "_id",
        foreignField: "_id",
        as: "restaurant"
    }
},
{
    $unwind: "$restaurant"
},
{
    $project: {
        _id:false,
        restaurant_id:"$_id",
        categories:true,
        restaurant_data:"$restaurant"
    }
}
]);