Question

我需要从数据库中检索整个单个对象层次结构作为JSON。实际上关于实现这一结果的任何其他解决方案的提议将受到高度关注。我决定使用MongoDB及其$ lookup支持。

所以我有三个系列：

方

{ "_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" }

addressComment

{ "_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" }

我需要检索具有所有相应地址的所有方并将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"
    }
}])

结果非常奇怪。有些记录还可以。但是没有_id 4的派对（没有地址）。结果集中还有两个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很陌生，但我觉得它可以做我需要的东西。

Answer 1

你的烦恼的原因＆＃39;是第二个聚合阶段 - { $unwind: "$address" }。它会删除具有_id: 4的聚会记录（因为它的地址数组为空，如您所述）并为聚会_id: 1和_id: 5生成两条记录（因为它们中的每一个都有两个地址）。

为防止删除没有地址的参与方，您应将$unwind阶段的preserveNullAndEmptyArrays选项设置为true。
为了防止对不同地址的各方进行重复，您应该在管道中添加$group聚合阶段。此外，使用$project阶段与$filter运算符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

使用mongodb 3.6 及更高版本的$lookup语法，无需使用$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" }
])

MongoDB嵌套查找3级

2 个答案: