[英]C# mongodb with strongly typed ID using records
我正在使用帶有 Mongo 的 C#9 記錄來處理強類型 ID。 我已經成功地完成了插入工作,但讀取時拋出了異常(見下文)。
這些是 model 定義。 非常簡單, TestModel
有一個類型為StronglyTypedId<Guid>
的TestModelId
。
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;
這些需要注冊序列化提供程序和序列化程序 class。
登記:
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>)!;
}
}
下面用於插入新文檔的代碼有效。
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
讀取會引發異常: 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();
我嘗試使用Equals
代替,像這樣
TestModel found = collection.Find(x => x.Id.Equals(model.Id)).FirstOrDefault();
但這引發了另一個異常: 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]''
我不確定那里發生了什么,類型定義很清楚,但由於某種原因似乎類型不匹配? 我試圖用谷歌搜索任何解決方案,但找不到任何東西。
我沒有找到導致異常的原因,但我找到了解決方法:
var filter = Builders<TestModel>.Filter.Eq("_id", model.Id.Id);
TestModel found = collection.Find(filter).FirstOrDefault();
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