如何使用 MongoDB 将 `$lookup` 聚合为 `findOne()`
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How to use MongoDBs aggregate `$lookup` as `findOne()`
提问by Fizzix
So as you all know, find()returns an array of results, with findOne()returning just a simply object.
众所周知，find()返回一个结果数组，findOne()只返回一个简单的对象。
With Angular, this makes a huge difference. Instead of going {{myresult[0].name}}, I can simply just write {{myresult.name}}.
使用 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}}。
I have found that the $lookupmethod in the aggregate pipeline returns an array of results instead of just a single object.
我发现$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。
For example, I have two colletions:
例如，我有两个集合：
userscollection:
users收藏：
[{
"firstName": "John",
"lastName": "Smith",
"country": 123
}, {
"firstName": "Luke",
"lastName": "Jones",
"country": 321
}]
countriescollection:
countries收藏：
[{
"name": "Australia",
"code": "AU",
"_id": 123
}, {
"name": "New Zealand",
"code": "NZ",
"_id": 321
}]
My aggregate $lookup:
我的聚合$lookup：
db.users.aggregate([{
$project: {
"fullName": {
$concat: ["$firstName", " ", "$lastName"]
},
"country": "$country"
}
}, {
$lookup: {
from: "countries",
localField: "country",
foreignField: "_id",
as: "country"
}
}])
The results from the query:
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
}]
}]
As you can see by the above results, each countryis an array instead of a single object like "country": {....}.
从上面的结果可以看出，每个country都是一个数组，而不是像"country": {....}.
How can I have my $lookupreturn a single object instead of an array since it will only ever match a single document?
我怎样才能让我$lookup返回单个对象而不是数组，因为它只会匹配单个文档？
回答by styvane
You're almost there, you need to add another $projectstage to your pipeline and use the $arrayElemAtto return the single element in the array.
您快完成了，您需要向$project管道添加另一个阶段并使用$arrayElemAt返回数组中的单个元素。
db.users.aggregate(
[
{ "$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 ] }
}}
]
)
回答by Ros
You can also use "preserveNullAndEmptyArrays"
你也可以使用 "preserveNullAndEmptyArrays"
Like so:
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
}
},
]
)
