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.我决定使用具有$ lookup支持的MongoDB.
I need to retrieve the entire single object hierarchy from the database as a JSON. Actually the proposal about any other solution to achive this result would be highly appriciated. I decided to use MongoDB with its $lookup support.
所以我有三个收藏夹:
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" }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:
I need to retrieve all parties with all corresponding addresses and address comments as part of the record. My aggregation:
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):
The result is pretty weird. Some records are ok. But Party with _id 4 is missing (there is no address for it). Also there are two Party _id 1 in the result set (but with different addresses):
{ "_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" }] } }请帮助我.我对MongoDB相当陌生,但是我觉得它可以满足我的需求.
Please help me with this. I'm pretty new to the MongoDB but I feel it can do what I need from it.
推荐答案麻烦"的原因是第二个汇总阶段-{ $unwind: "$address" }.它将删除具有_id: 4的一方的记录(因为您的地址数组为空,如您所提到的),并为具有_id: 1和_id: 5的一方生成两条记录(因为它们每个都有两个地址).
The cause of your 'troubles' is the second aggregation stage - { $unwind: "$address" }. It removes record for party with _id: 4 (because its address array is empty, as you mention) and produces two records for parties _id: 1 and _id: 5 (because each of them has two addresses).
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为防止删除无地址的参与者,您应该设置preserveNullAndEmptyArrays选项noreferrer> $unwind 过渡到true.
To prevent removing of parties without addresses you should set preserveNullAndEmptyArrays option of $unwind stage to true.
为防止当事方重复其不同地址,您应该添加 $group 聚集到您的管道中.另外,将 $project 阶段与 $filter 运算符可排除输出中的空地址记录.
To prevent duplicating of parties for its different addresses you should add $group aggregation stage to your pipeline. Also, use $project stage with $filter operator to exclude empty address records in output.
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] } } } } }]);更多推荐
具有3个级别的MongoDB嵌套查找
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