c# - 通过 WCF 抛出异常的正确方法

Question

我试图以最通用的方式通过 WCF 发送异常。这是我得到的:

[ServiceContract]
interface IContract
{
    [OperationContract]
    void Foo();
}

class ContractImplementation: IContract
{
    public void Foo()
    {
        try
        {
            Bar();
        }
        catch (Exception ex)
        {
            throw new FaultException<Exception>(ex, ex.Message);
        }
    }
}

Bar 中实际出现的异常是:

[Serializable]
class MyException : Exception
{
    // serialization constructors
}

我在服务器端 WCF 日志记录中看到的错误是:

Type 'MyException' with data contract name 'MyException:http://schemas.datacontract.org/2004/07/MyException' is not expected. Consider using a DataContractResolver or add any types not known statically to the list of known types - for example, by using the KnownTypeAttribute attribute or by adding them to the list of known types passed to DataContractSerializer.

到目前为止我尝试了什么:

[ServiceKnownType(typeof(MyException))]
[ServiceContract]
interface IContract
{
    [FaultContract(typeof(MyException))]
    [OperationContract]
    void Foo();
}

但没有运气。

Answer 1

首先，在 MyException 中，移除对 Exception 的继承并将其公开。

其次，当你声明你的服务契约(Contract)时，声明异常如下:

[FaultContractAttribute(
        typeof(MyException),
        Action = "", 
        Name = "MyException", 
        Namespace = "YourNamespace")]
    [System.ServiceModel.XmlSerializerFormatAttribute(SupportFaults = true)]
    [OperationContract]
    void Foo()

最后，您可以像这样抛出异常:

throw new FaultException<MyException>
             (
                 new MyException(ex.Message),
                 new FaultReason("Description of your Fault")

             );

希望对您有所帮助。