如何在引发异常之前将错误消息传递回去

Question

>我有以下代码收到错误消息.我想在抛出异常之前将其传递到字符串中，这是我的代码

ValidateError(authDeserialized, "Succeed", "error", "failed"); //the validateError is a function as indicated below
Model.Response= authResponse.Content;
protected static void ValidateError(dynamic response, string validStatus,string categoryMatch, string message)
{  
if (response.result.status != validStatus)
{
try
{
var category = response.result.category;
if (category == categoryMatch)
message=ErrorCodes.MessageFor(code,description);
//so i get the message back fine here but now how do i pass it back to this line   Model.Response= authResponse.Content; so that it can get saved?
}
catch (Exception) { }
throw new Exception(message ?? "Request was not successfull");
}
}

Answer 1

由于您已经将message发送到ValidateError()方法，请将该参数作为out参数传递，它将更新message的值，如果您为其分配新值，它将更新消息并可供外部环境访问。

string failureMessage = "failed";
ValidateError(authDeserialized, "Succeed", "error", out failureMessage);
//^^^ This is what you have to change
//Now you can assign failureMessage to any other value
Model.Response= authResponse.Content;
protected static void ValidateError(dynamic response, string validStatus,string categoryMatch, out string message)
{                                                                                            //^^^ This is what you have to change
if (response.result.status != validStatus)
{
try
{
var category = response.result.category;
if (category == categoryMatch)
message=ErrorCodes.MessageFor(code,description);    //so i get the message back fine here but now how do i pass it back to this line   Model.Response= authResponse.Content; so that it can get saved?
}
catch (Exception) { }
throw new Exception(message ?? "Request was not successfull");
}
}

通过这种方式，您可以在引发错误之前为失败消息赋值。

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