mongodb - 如何使用 MongoDB 将 `$lookup` 聚合为 `findOne()`

Question

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

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

我发现 $lookup聚合管道中的方法返回一个结果数组，而不仅仅是一个对象。

例如，我有两个集合:

users收藏:

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

countries收藏:

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

我的汇总 $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
 }]
}]

从上面的结果可以看出，每个 country是一个数组，而不是像 "country": {....} 这样的单个对象.

我怎样才能拥有我的$lookup返回单个对象而不是数组，因为它只会匹配单个文档？

Answer 1

您快到了，您需要添加另一个 $project转移到您的管道并使用 $arrayElemAt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 ] } 
        }} 
    ]
)