如何使用MongoDB将`$ lookup`聚合为`findOne()`

Question

众所周知,find()返回一个结果数组,findOne()只返回一个简单的对象.

使用Angular,这会产生巨大的差异.{{myresult[0].name}}我可以简单地写一下,而不是去{{myresult.name}}.

我发现$lookup聚合管道中的方法返回结果数组而不是单个对象.

例如,我有两个收藏:

users 采集:

[{
  "firstName": "John",
  "lastName": "Smith",
  "country": 123
}, {
  "firstName": "Luke",
  "lastName": "Jones",
  "country": 321
}]

countries 采集:

[{
  "name": "Australia",
  "code": "AU",
  "_id": 123
}, {
  "name": "New Zealand",
  "code": "NZ",
  "_id": 321
}]

我的总计$lookup:

db.users.aggregate([{
  $project: {
    "fullName": {
      $concat: ["$firstName", " ", "$lastName"]
    },
    "country": "$country"
  }
}, {
  $lookup: {
    from: "countries",
    localField: "country",
    foreignField: "_id",
    as: "country"
  }
}])

查询结果:

[{
  "fullName": "John Smith",
  "country": [{
    "name": "Australia",
    "code": "AU",
    "_id": 123
  }]
}, {
 "fullName": "Luke Jones",
 "country": [{
   "name": "New Zealand",
   "code": "NZ",
   "_id": 321
 }]
}]

正如您在上面的结果中所看到的,每个country都是一个数组,而不是像一个对象"country": {....}.

如何让我$lookup返回单个对象而不是数组,因为它只能匹配单个文档？

Answer 1

你几乎就在那里,你需要$project在管道中添加另一个阶段并使用它$arrayElemAt来返回数组中的单个元素.

db.users.aggregate(
    [
        {   "$project": {     
            "fullName": {       
                "$concat": [ "$firstName", " ", "$lastName"]     
            },
            "country": "$country"   
        }}, 
        { "$lookup": {     
                "from": "countries",     
                "localField": "country",     
                "foreignField": "_id",     
                "as": "countryInfo"   
        }}, 
        { "$project": { 
            "fullName": 1, 
            "country": 1, 
            "countryInfo": { "$arrayElemAt": [ "$countryInfo", 0 ] } 
        }} 
    ]
)

Answer 2

db.users.aggregate([
    {
        $lookup: {
            from: 'countries',
            localField: 'country',
            foreignField: '_id',
            as: 'country'
        }
    },
    {
        $unwind: '$country'
    }
]).pretty()

您可以使用此 mongo 查询来获取国家/地区对象

Answer 3

您也可以使用 "preserveNullAndEmptyArrays"

像这样：

 db.users.aggregate(
        [
            {   "$project": {     
                "fullName": {       
                    "$concat": [ "$firstName", " ", "$lastName"]     
                },
                "country": "$country"   
            }}, 
            { "$lookup": {     
                    "from": "countries",     
                    "localField": "country",     
                    "foreignField": "_id",     
                    "as": "countryInfo"   
            }}, 
            {"$unwind": {
                    "path": "$countryInfo",
                    "preserveNullAndEmptyArrays": true
                }
            },
        ]
    )

Answer 4

当您不想重复项目中的所有字段时，只需使用 $addFields 覆盖有问题的字段：

db.users.aggregate([
    {   "$project": {     
        "fullName": {       
            "$concat": [ "$firstName", " ", "$lastName"]     
        },
        "country": "$country"   
    }}, 
    { "$lookup": {     
        "from": "countries",     
        "localField": "country",     
        "foreignField": "_id",     
        "as": "countryInfo"   
    }},
    { "$addFields": {
        "countryInfo": {
            "$arrayElemAt": [ "$countryInfo", 0 ]
        }
    }}
])