具有3个级别的MongoDB嵌套查找

Yuriy 发表于 Dev

尤里

我需要从数据库中以JSON检索整个单个对象层次结构。实际上，关于实现此结果的任何其他解决方案的建议将被高度推荐。我决定使用MongoDB及其$ lookup支持。

因此，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" }

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" }

我需要检索所有具有所有相应地址和地址注释的各方，作为记录的一部分。我的汇总：

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。某些记录还可以。但是缺少_id 4的Party（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"
        }]
    }
}

请帮我解决一下这个。我对MongoDB还是很陌生，但是我觉得它可以满足我的需求。

您“麻烦”的原因是第二个汇总阶段- { $unwind: "$address" }。它删除带有的参与者的记录_id: 4（因为，如前所述，因为其地址数组为空），并为产生两个记录，_id: 1并且_id: 5（因为每个记录都有两个地址）。

为了防止没有地址的一方被除名，您应该将stagepreserveNullAndEmptyArrays选项设置$unwind为true。
为了防止各方为其不同的地址重复，您应该$group在管道中添加聚合阶段。另外，使用$project带有$filter运算符的stage可以在输出中排除空地址记录。

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] } }
    } 
  }
}]);