MongoDB 3 级嵌套查找

Question

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，我们将高度赞赏有关实现此结果的任何其他解决方案的提案。我决定使用 MongoDB 及其 $lookup 支持。

所以我有三个合集：

聚会

{ "_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" }

我需要检索所有相关方的所有相应地址和地址注释作为记录的一部分。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。有些记录还可以。但是 _id: 4 的派对不见了（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。我是 MongoDB 的新手，但我觉得它可以满足我的需求。

Answer 1

您 'troubles' 的原因是第二个聚合阶段 - { $unwind: "$address" }。它删除了 _id: 4 方的记录（因为它的地址数组是空的，正如你提到的）并为方 _id: 1 和 _id: 5 生成两个记录（因为它们每个都有两个地址）。

为了防止移除没有地址的派对，您应该将 $unwind 阶段的 preserveNullAndEmptyArrays 选项设置为 true。
为了防止不同地址的各方重复，您应该添加 $group aggregation stage to your pipeline. Also, use $project stage with $filter 运算符以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] } }
    } 
  }
}]);

Answer 2

同mongodb3.6及以上$lookup syntax it is quite simple to join nested fields without using $unwind.

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" }
])

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