如何使用 Mongodb Left join 以及条件获得所有结果？

Question

我有两张桌子。 1) 房间列表 2) 房间用户。这里的左表将是 roomList。即使连接的表结果为空，是否可以获取所有结果？

我目前正在使用聚合：$lookup、$unwind、$match。

房间列表

const roomInfo = new mongoose.Schema({
    roomName: {
        type: String,
        trim: true,
        required: true
    },
    currentMembers: {
        type: Number,
        required: true
    },
    createdBy: {
        type: mongoose.Schema.ObjectId,
        required: true,
        ref: "User"
    },
    creationDate: {
        type: Date,
        trim: true
    },
    status: {
        type: Number,
        required: true,
        default: 1
    }
});

房间用户

const roomUser = new mongoose.Schema({
    roomId: {
        type: mongoose.Schema.ObjectId,
        required: true,
        ref: "RoomInfo"
    },
    userId: {
        type: mongoose.Schema.ObjectId,
        required: true,
        ref: "User"
    },
    joiningDate: {
        type: Date,
        trim: true
    },
    leavingDate: {
        type: Date,
        trim: true
    },
    requestDate:{
        type:Date,
        trim:true
    },
    status: {
        type: Number,
        default: 0 
    }

我想要类似于这个 mysql 查询的结果。

SELECT * FROM roomList LEFT JOIN roomUser ON roomUser.roomId = roomList.id WHERE roomUser.userId = 123 AND roomList.status = 1。

Answer 1

它可以通过多个聚合阶段来完成。

1. $match by roomUser id
2. $lookup by room_id
3. $unwind 找到的房间
4. $匹配状态
5. $replaceRoot 推广房间

db.roomUser.aggregate([
    {
        "$match": {
            "_id": ObjectId("5d3473a686041613147424ad")
        }
    },
    {
        "$lookup": {
            "from": "roomList",
            "localField": "room_id",
            "foreignField": "_id",
            "as": "r"
        }
    },
    {
        "$unwind": "$r"
    },
    {
        "$match": {
            "r.status": 1
        }
    },
    {
        "$replaceRoot": {
            "newRoot": "$r"
        }
    }
])

我不知道猫鼬，希望你能想出猫鼬等效...

Answer 2

我终于明白了，使用所谓的 JOIN 来解决这个问题太难了（撞头）。我使用猫鼬子文档做到了。现在我的收藏看起来像这样：

const mongoose = require('mongoose');
const Schema = mongoose.Schema;
const roomUsers = new Schema (
    {
        userId: {
            type: mongoose.Schema.ObjectId,
            required: true,
            ref: "User"
        },
        joiningDate: {
            type: Date,
            trim: true
        },
        leavingDate: {
            type: Date,
            trim: true
        },
        requestDate:{
            type:Date,
            trim:true
        },
        status: {
            type: Number,
            default: 0 // 0 for none, 1 for request, 2 for joined, 3 for blocked
        }
    }
);

const roomInfo = new mongoose.Schema({
    roomName: {
        type: String,
        trim: true,
        required: true
    },
    currentMembers: {
        type: Number,
        required: true
    },
    createdBy: {
        type: mongoose.Schema.ObjectId,
        required: true,
        ref: "User"
    },
    creationDate: {
        type: Date,
        trim: true
    },
    status: {
        type: Number,
        required: true,
        default: 1 // 1 for active, 2 for deactive
    },
    roomUsers:[roomUsers]
});

var roomInfp = mongoose.model("RoomInfo", roomInfo);
module.exports = roomInfp;

Answer 3

MongoDb 和 noSql 数据库的理念与 SQL 数据库不同。

对于 SQL 数据库，我们应该创建两个表来存储 roomlist 一个 roomUser。 在 NoSQL 数据库中，您应该将用户数据存储在房间列表中、特殊条目（数组、新对象）中，或者创建一个新表来存储用户和房间列表之间的链接。

之后，您应该可以进行查询了。

最好的问候