具有 3 个级别的 MongoDB 嵌套查找

Posted 2023-02-16

【中文标题】具有 3 个级别的 MongoDB 嵌套查找

【英文标题】：MongoDB nested lookup with 3 levels

【发布时间】：2016-07-01 09:22:24

【问题描述】：

我需要从数据库中检索整个单个对象层次结构作为 JSON。实际上，将高度赞赏有关实现此结果的任何其他解决方案的建议。我决定使用带有 $lookup 支持的 MongoDB。

所以我有三个集合：

派对

 "_id" : "2", "name" : "party2" 
 "_id" : "5", "name" : "party5" 
 "_id" : "4", "name" : "party4" 
 "_id" : "1", "name" : "party1" 
 "_id" : "3", "name" : "party3"     

地址

 "_id" : "a3", "street" : "Address3", "party_id" : "2" 
 "_id" : "a6", "street" : "Address6", "party_id" : "5" 
 "_id" : "a1", "street" : "Address1", "party_id" : "1" 
 "_id" : "a5", "street" : "Address5", "party_id" : "5" 
 "_id" : "a2", "street" : "Address2", "party_id" : "1" 
 "_id" : "a4", "street" : "Address4", "party_id" : "3" 

地址评论

 "_id" : "ac2", "address_id" : "a1", "comment" : "Comment2" 
 "_id" : "ac1", "address_id" : "a1", "comment" : "Comment1" 
 "_id" : "ac5", "address_id" : "a5", "comment" : "Comment6" 
 "_id" : "ac4", "address_id" : "a3", "comment" : "Comment4" 
 "_id" : "ac3", "address_id" : "a2", "comment" : "Comment3" 

我需要检索所有具有所有相应地址的各方，并将地址 cmet 作为记录的一部分。我的聚合：

db.party.aggregate([
    $lookup: 
        from: "address",
        localField: "_id",
        foreignField: "party_id",
        as: "address"
    
,

    $unwind: "$address"
,

    $lookup: 
        from: "addressComment",
        localField: "address._id",
        foreignField: "address_id",
        as: "address.addressComment"
    
])

结果很奇怪。有些记录没问题。但是缺少_id: 4 的派对（没有地址）。另外，结果集中有两个Party_id: 1（但地址不同）：

    "_id": "1",
    "name": "party1",
    "address": 
        "_id": "2",
        "street": "Address2",
        "party_id": "1",
        "addressComment": [
            "_id": "3",
            "address_id": "2",
            "comment": "Comment3"
        ]
    

    "_id": "1",
    "name": "party1",
    "address": 
        "_id": "1",
        "street": "Address1",
        "party_id": "1",
        "addressComment": [
            "_id": "1",
            "address_id": "1",
            "comment": "Comment1"
        ,
        
            "_id": "2",
            "address_id": "1",
            "comment": "Comment2"
        ]
    

    "_id": "3",
    "name": "party3",
    "address": 
        "_id": "4",
        "street": "Address4",
        "party_id": "3",
        "addressComment": []
    

    "_id": "5",
    "name": "party5",
    "address": 
        "_id": "5",
        "street": "Address5",
        "party_id": "5",
        "addressComment": [
            "_id": "5",
            "address_id": "5",
            "comment": "Comment5"
        ]
    

    "_id": "2",
    "name": "party2",
    "address": 
        "_id": "3",
        "street": "Address3",
        "party_id": "2",
        "addressComment": [
            "_id": "4",
            "address_id": "3",
            "comment": "Comment4"
        ]
    

请帮我解决这个问题。我对 MongoDB 还很陌生，但我觉得它可以满足我的需求。

【参考方案1】：

“麻烦”的原因是第二个聚合阶段 - $unwind: "$address" 。它删除了_id: 4 的记录（因为它的地址数组是空的，正如你提到的）并为_id: 1 和_id: 5 的当事人生成两条记录（因为他们每个人都有两个地址）。

为防止删除没有地址的各方，您应将$unwind 阶段的preserveNullAndEmptyArrays 选项设置为true。

为了防止各方重复其不同的地址，您应该将$group 聚合阶段添加到您的管道。另外，使用$project 阶段和$filter 运算符来排除输出中的空地址记录。

db.party.aggregate([
  $lookup: 
    from: "address",
    localField: "_id",
    foreignField: "party_id",
    as: "address"
  
, 
  $unwind: 
    path: "$address",
    preserveNullAndEmptyArrays: true
  
, 
  $lookup: 
    from: "addressComment",
    localField: "address._id",
    foreignField: "address_id",
    as: "address.addressComment",
  
, 
  $group: 
    _id : "$_id",
    name:  $first: "$name" ,
    address:  $push: "$address" 
  
, 
  $project: 
    _id: 1,
    name: 1,
    address: 
      $filter:  input: "$address", as: "a", cond:  $ifNull: ["$$a._id", false]  
     
  
]);

【讨论】：

坦克你鲥鱼！但是记录 4 有一个小问题： "_id": "4", "name": "party4", "address": [ "addressComment": [] ] 如您所见 - 地址应该为空，但它是一个空记录...如果地址注释为空，我们可以跳过地址吗？在其他情况下，此地址将被视为记录。

实际上，根据 $unwind 操作的新“preserveNullAndEmptyArrays”字段的描述（自 3.2 起），我看到提供的解决方案按预期工作。现在我们可以跳过“$project”这一步并指定这个“$unwind”而不是简单的一个：$unwind: path: "$address",preserveNullAndEmptyArrays: true。我会接受你的回答，感谢你快速而清晰的回复！

@Shad 我有非常相似的问题。这里 OP 的代码在party 集合中只有一个名为name 的属性，因此您使用$first 在$group 中获取它。假设我有 10 多个属性，那么有什么方法可以自动获取所有属性而不单独提及每个属性？

如果同一地址 id 有不同的 cmets，我们如何为“addressComment”添加组子句？如前所述，此查询适用于“地址”级别的分组。

【参考方案2】：

使用 mongodb 3.6 及以上$lookup 语法，无需使用$unwind 即可连接嵌套字段非常简单。

db.party.aggregate([
   "$lookup": 
    "from": "address",
    "let":  "partyId": "$_id" ,
    "pipeline": [
       "$match":  "$expr":  "$eq": ["$party_id", "$$partyId"] ,
       "$lookup": 
        "from": "addressComment",
        "let":  "addressId": "$_id" ,
        "pipeline": [
           "$match":  "$expr":  "$eq": ["$address_id", "$$addressId"] 
        ],
        "as": "address"
      
    ],
    "as": "address"
  ,
   "$unwind": "$address" 
])

【讨论】：

如果需要，您将如何访问内部管道中的 partyId？

@Paulo 使用$$ 符号

你确定吗，我知道你可以像这样得到addressId $$addressId

@Paulo 请参考this link in the mongo docs 了解清楚

嗨@Ashh，你能在***.com/questions/70021771/…查看我的问题