top*_*olm 1 c# mongodb-.net-driver
如果我有对象,让它们称为Group,其中包含一些其他对象的列表,我将其称为Brand,并且该对象具有一个名为Model的对象列表。
有没有办法只获取使用 MongoDb c# 驱动程序的模型列表。
我多次尝试使用SelectMany但没有成功。如果我放了多个SelectMany,我总是得到一个空列表。
代码应该是不言自明的。最后是解释什么让我感到困惑的评论。
class Group
{
[BsonId(IdGenerator = typeof(GuidGenerator))]
public Guid ID { get; set; }
public string Name { get; set; }
public List<Brand> Brands { get; set; }
}
class Brand
{
public string Name { get; set; }
public List<Model> Models { get; set; }
}
class Model
{
public string Name { get; set; }
public int Produced { get; set; }
}
class Program
{
static void Main(string[] args)
{
MongoClient client = new MongoClient("mongodb://127.0.0.1:32768");
var db = client.GetDatabase("test");
var collection = db.GetCollection<Group>("groups");
var fca = new Group { Name = "FCA" };
var alfaRomeo = new Brand { Name = "Alfra Romeo" };
var jeep = new Brand { Name = "Jeep" };
var ferrari = new Brand { Name = "Ferrari"};
var laFerrari = new Model { Name = "LaFerrari", Produced = 4 };
var wrangler = new Model { Name = "Wrangler", Produced = 3 };
var compass = new Model { Name = "Compass", Produced = 8 };
var giulietta = new Model { Name = "Giulietta", Produced = 7 };
var giulia = new Model { Name = "Giulia", Produced = 8 };
var _4c = new Model { Name = "4C", Produced = 6 };
fca.Brands = new List<Brand> { ferrari, alfaRomeo, jeep };
ferrari.Models = new List<Model> { laFerrari };
jeep.Models = new List<Model> { wrangler, compass };
alfaRomeo.Models = new List<Model> { giulietta, giulia, _4c };
collection.InsertOne(fca);
Console.WriteLine("press enter to continue");
Console.ReadLine();
var models = collection.AsQueryable().SelectMany(g => g.Brands).SelectMany(b => b.Models).ToList();
Console.WriteLine(models.Count); //returns 0 I expected 6
Console.ReadLine();
}
}
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尝试
var models = collection.AsQueryable()
.SelectMany(g => g.Brands)
.Select(y => y.Models)
.SelectMany(x=> x);
Console.WriteLine(models.Count());
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工作输出(带有额外的 Select())
aggregate([{
"$unwind": "$Brands"
}, {
"$project": {
"Brands": "$Brands",
"_id": 0
}
}, {
"$project": {
"Models": "$Brands.Models",
"_id": 0
}
}, {
"$unwind": "$Models"
}, {
"$project": {
"Models": "$Models",
"_id": 0
}
}])
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没有额外 Select() 的 OP 输出
aggregate([{
"$unwind": "$Brands"
}, {
"$project": {
"Brands": "$Brands",
"_id": 0
}
}, {
"$unwind": "$Models"
}, {
"$project": {
"Models": "$Models",
"_id": 0
}
}])
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