WCF:许多方法上的相同故障契约(Contract)

Question

以一个有 10 个服务和每个服务 20 个方法的项目为例。

所有服务都继承自具有安全检查的基础服务。每个方法所做的第一件事就是调用安全检查。如果出现问题，这将引发安全异常。

问题是:我需要在每个方法 (OperationContract) 上指定一个 FaultContract，还是可以在中央定义中指定一次？

Answer 1

您可以通过创建自定义属性来实现。

实现 IContractBehavior 并将错误添加到 Validate 方法上的每个操作。

void IContractBehavior.Validate(ContractDescription contractDescription, ServiceEndpoint endpoint)
{
   foreach (OperationDescription od in contractDescription.Operations)
      od.Add(yourFault);
}

这是a link详细说明了如何实现这一点。下面是要使用的实际代码:

[AttributeUsage(AttributeTargets.Interface, AllowMultiple = false, Inherited = true)]
public class StandardFaultsAttribute : Attribute, IContractBehavior
{
    // this is a list of our standard fault detail classes.
    static Type[] Faults = new Type[]
    {
        typeof(AuthFailure),
        typeof(UnexpectedException),
        typeof(UserFriendlyError)
    };

    public void AddBindingParameters(ContractDescription contractDescription, ServiceEndpoint endpoint, BindingParameterCollection bindingParameters)
    {
    }

    public void ApplyClientBehavior(ContractDescription contractDescription, ServiceEndpoint endpoint, ClientRuntime clientRuntime)
    {
    }

    public void ApplyDispatchBehavior(ContractDescription contractDescription, ServiceEndpoint endpoint, DispatchRuntime dispatchRuntime)
    {
    }

    public void Validate(ContractDescription contractDescription, ServiceEndpoint endpoint)
    {
        foreach (OperationDescription op in contractDescription.Operations)
        {
            foreach (Type fault in Faults)
            {
                op.Faults.Add(MakeFault(fault));
            }
        }
    }

    private FaultDescription MakeFault(Type detailType)
    {
        string action = detailType.Name;
        DescriptionAttribute description = (DescriptionAttribute)                Attribute.GetCustomAttribute(detailType, typeof(DescriptionAttribute));

        if (description != null)
            action = description.Description;
        FaultDescription fd = new FaultDescription(action);
        fd.DetailType = detailType;
        fd.Name = detailType.Name;
        return fd;
    }
}