C# mongodb 使用记录的强类型 ID

Question

我正在使用带有 Mongo 的 C#9 记录来处理强类型 ID。我已经成功地进行了插入工作，但读取时抛出异常（见下文）。

这些是模型定义。很简单，TestModel 有一个 TestModelId 类型为 StronglyTypedId<Guid>.

public record TestModel(TestModelId Id, string value, string? value1);
public record TestModelId(Guid Id) : StronglyTypedId<Guid>(Id);
public abstract record StronglyTypedId<TValue>(TValue Id) where TValue : notnull;

这些需要注册序列化提供程序和序列化程序 class。

注册：

var guidSerializer = BsonSerializer.SerializerRegistry.GetSerializer<Guid>();
BsonSerializer.RegisterSerializationProvider(new IdSerializationProvider(guidSerializer));

序列化程序：

public class IdSerializationProvider : IBsonSerializationProvider
{
    private IBsonSerializer<Guid> guidSerializer;

    public IdSerializationProvider(IBsonSerializer<Guid> guidSerializer)
    {
        this.guidSerializer = guidSerializer;
    }

    public IBsonSerializer? GetSerializer(Type type)
    {
        if(type.BaseType != null && type.BaseType == typeof(StronglyTypedId<Guid>)) 
            return new TestModelSerializer(type, guidSerializer);
        
        return null;
    }
}

public class TestModelSerializer : SerializerBase<StronglyTypedId<Guid>>
{
    private readonly Type targetType;
    private readonly IBsonSerializer<Guid> guidSerializer;

    public TestModelSerializer(Type targetType, IBsonSerializer<Guid> guidSerializer)
    {
        this.targetType = targetType;
        this.guidSerializer = guidSerializer;
    }

    public override void Serialize(BsonSerializationContext context, BsonSerializationArgs args, StronglyTypedId<Guid> value)
    {
        guidSerializer.Serialize(context, args, value.Id);
    }

    public override StronglyTypedId<Guid> Deserialize(BsonDeserializationContext context, BsonDeserializationArgs args)
    {
        var guid = guidSerializer.Deserialize(context, args);
        var idType = Activator.CreateInstance(targetType, new object[] { guid });
        return (idType as StronglyTypedId<Guid>)!;
    }
}

下面用于插入新文档的代码有效。

MongoClient dbClient = new MongoClient("mongodb://localhost:27017");
IMongoCollection<TestModel> collection = dbClient.GetDatabase("test").GetCollection<TestModel>("testmodel");

TestModel model = new TestModel(new TestModelId(Guid.NewGuid()), "test", "test");
collection.InsertOne(model);

但是使用 Find 读取会抛出异常： System.InvalidOperationException: 'The operands for operator 'Equal' do not match the parameters of method 'op_Equality'.'

TestModel found = collection.Find(x => x.Id == model.Id).FirstOrDefault();

我试过使用 Equals 代替，像这样

TestModel found = collection.Find(x => x.Id.Equals(model.Id)).FirstOrDefault();

但这是抛出另一个异常：System.ArgumentException: 'Method 'Boolean Equals(WebApplication2.Controllers.TestModelId)' declared on type 'WebApplication2.Controllers.TestModelId' cannot be called with instance of type 'WebApplication2.Controllers.StronglyTypedId`1[System.Guid]''

我不确定那里发生了什么，类型定义很清楚，但由于某种原因似乎类型不匹配？我尝试 google 任何解决方案，但找不到任何东西。

Answer 1

我没有找到导致异常的原因，但我找到了解决方法：

var filter = Builders<TestModel>.Filter.Eq("_id", model.Id.Id);
TestModel found = collection.Find(filter).FirstOrDefault();

C# mongodb 使用记录的强类型 ID

C# mongodb with strongly typed ID using records

c#

mongodb

c#-9.0