MongoDB 让 $lookup 只返回一条记录

Question

这是预订表

booking [
    {
        "_id" : "0J0DR",
        "user" : "MN90L",
        "property" : "58669471869659d70b424ea7",
        "checkin" : 1488758400,
        "checkout" : 1489363200
    },
    {
        "_id" : "0PDLR",
        "user" : "7CSEF",
        "property" : "586694ea869659d70b424eb3",
        "checkin" : 1488326400,
        "checkout" : 1498780800
    }
]

这是用户表

users [
    {
        "_id" : "4M4KE",
        "email" : "test@vest.com",
        "name" : "Torben"
    },
    {
        "_id" : "MN90L",
        "email" : "mr@booker.com",
        "name" : "Mr. Booker"
    },
    {
        "_id" : "GF37A",
        "email" : "test@test.com",
        "name" : "Whatever"
    },
    {
        "_id" : "7CSEF",
        "email" : "miss@booker.com",
        "name" : "Miss. Booker"
    },
    {
        "_id" : "W0LG9",
        "email" : "xxx@yyy.com",
        "name" : "Whatever"
    }
]

这是我的查询，效果很好，除了每个预订记录都附加了所有用户，而不仅仅是用户。_id = booking.user

db.getCollection('booking').aggregate([{
    $match: {
        checkin : {$lte: (1512145439)},
        checkout: {$gte: (1483203600)},
    }
}, {
    $lookup: {
        from: "users",
        localField: "users._id",
        foreignField: "booking.user",
        as: "users"
    }   
}, {
    $unwind: "$users"
}])

我做错了什么？

我尝试了使用和不使用 $unwind。

我需要做一些 $match 还是 localField foreignField 就足够了？

我的 db.version() = 3.2.12

Answer 1

您混淆了 localField 和 foreignField。

尝试不带别名/集合名称的 localField: "user", foreignField: "_id"。

来自文档

localField: field from the input documents,
foreignField: field from the documents of the "from" collection,

https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/

当您使用点符号时，MongoDB 认为您正在尝试访问嵌入式文档中的字段。

https://docs.mongodb.com/manual/core/document/#document-dot-notation