MongoDB条件关系查询

如何解决MongoDB条件关系查询

MongoDB 4.2.2，

我有2个投票和投票集合，我需要从“投票”集合中抓取字段voter_selection并将其添加到投票中，但前提是某些条件匹配poll_id和voter_id >

用户：

{
"_id" : ObjectId("5f867e0d126ddbde24d6ee73"),"name" : "Jaskson"
"age" : "24"
}

投票：

{
"_id" : ObjectId("5f87d988ddae726a3dbe5459"),"name" : "RedVsWhite"
},{
"_id" : ObjectId("5f51408ffc1d0437fa31d6f7"),"name" : "ApplesVsOrange","total_votes" : "0",}

投票：

{
"_id" : ObjectId("5f864addddae726a3dbe53de"),"voter_id" : ObjectId("5f867e0d126ddbde24d6ee73"),//this is the id of the person
"poll_id" : ObjectId("5f87d988ddae726a3dbe5459"),//this is the poll id
"voter_selection" : "red"
}

我需要这个结果：

{
"_id" : ObjectId("5f87d988ddae726a3dbe5459"),"name" : "RedVsWhite","voter_selection" : red // show this if it finds votes with this _id and also if user_id match & voter_id
},"total_votes" : "0"
/// dont show voter_selection if nothing match with voter_id and poll_id
}

此查询有效，但问题是，如果没有匹配项，则不会显示民意测验行，而我需要的是即使没有匹配项也要获得所有民意测验，但是如果有匹配项，请添加字段{{1} }移至该行

voter_selection

解决方法

• $group设置为空，并形成根所有元素的数组
• $lookup与VotesRecords加入收藏并获取用户详细信息
• $unwind解构VotesRecords数组
• $project迭代遍历$map数组上的root，并检查条件是否匹配poll_id，然后返回voter_selection，否则不使用
• $unwind解构根数组
• $replaceWith替换根中的root对象

db.getCollection("Polls").aggregate([
  {
    $group: {
      _id: null,root: { $push: "$$ROOT" }
    }
  },{
    $lookup: {
      from: "VotesRecords",as: "VotesRecords",pipeline: [{ $match: { voter_id: ObjectId("5f867e0d126ddbde24d6ee73") } }]
    }
  },{
    "$unwind": {
      "path": "$VotesRecords","preserveNullAndEmptyArrays": true
    }
  },{
    $project: {
      root: {
        $map: {
          input: "$root",in: {
            $mergeObjects: [
              "$$this",{
                $cond: [
                  { $eq: ["$$this._id","$VotesRecords.poll_id"] },{ voter_selection: "$VotesRecords.voter_selection" },{}
                ]
              }
            ]
          }
        }
      }
    }
  },{ $unwind: "$root" },{ $replaceWith: "$root" }
])

Playground

---

第二种方法，您可以在管道之后添加以下更改，

• $project显示voter_selection，如果voter_id匹配，否则显示$$REMOVE
• $group by _id，并使用$mergeObjects
从众多对象中获取唯一对象
• $replaceWith替换根对象

  // <= skipping your pipelines here
  {
    "$project": {
      "_id": "$_id","name": "$name","total_votes": 1,"voter_selection": {
        $cond: [
          { $eq: ["$VotesRecords.voter_id",ObjectId("5f867e0d126ddbde24d6ee73")] },"$VotesRecords.voter_selection","$$REMOVE"
        ]
      }
    }
  },{
    $group: {
      _id: "$_id",root: { $mergeObjects: "$$ROOT" }
    }
  },{ $replaceWith: "$root" }

Playground