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如何将格式良好的错误响应为JSON

asp.net-web-api json c#

xChaax · 技术社区 · 6 年前

嗨,我是C的新手,目前我正在构建我的第一个web API。要求将发生的任何错误以JSON格式显示为响应

这是我的答题课

public class ResponseData 
{
    public List<AccountDTO> AccountList { set; get; }
    public string ReturnMessage { set; get; }
    public string error { get; set; }
    public string code { get; set; }
    public string message { get; set; }
    public string innererror { get; set; }
    public string type { get; set; }
    public string stacktrace { get; set; }

}

这是我的控制器

例如,如果我的代码执行 catch 块,我将收集异常的所有信息并以格式化的格式返回 JSON

[HttpGet]
    public ResponseData getAllAccounts(string BRN, string CompanyName)
    {

        ResponseData response = new ResponseData();
        List<AccountDTO> accountlist = null;
        IAccountInterface usecase = new AccountDAO();
        try
        {
            if (BRN != null && CompanyName != null)
            {
                accountlist = usecase.getAccountbyBRNCompanyName().Where(p => p.BRN.ToLower().Equals(BRN.ToLower()) && p.CompanyName.ToLower().Contains(CompanyName.ToLower())).ToList();
                response.ReturnMessage = "Data filter by BRN and CompanyName.";
            }

            else
            {
                response.ReturnMessage = "Data filter cannot be null.";
            }
            response.AccountList = accountlist;
        }
        catch (Exception e)
        {
            Console.WriteLine("Error: " + e.Message + ", Stack: " + e.StackTrace);
            response.code = "sample code";
            response.message = e.Message;
            response.innererror = "{";
            response.message = e.Message;
            response.type = "ExceptionType + ...";
            response.stacktrace = e.StackTrace;



            //response.ReturnMessage = JsonConvert.SerializeObject(errorFormat);


        }

        return response;
    }

response object 所以我的所有对象属性都会显示出来。

这是我想要实现的一个示例响应。有关错误的详细信息作为JSON包含在响应中。错误将采用此格式。任何帮助和指导都非常感谢。

{  
 "error":{  
  "code": "<This code is not related to..>",  
  "message": "<A message describing the error..>",  
  "innererror": {  
   "message": "<A message describing the error..>",  
   "type": "Microsoft.Crm.CrmHttpException",  
   "stacktrace": "<Details from the server about where the 
                  error occurred>"  
    }  
   }  
}

2 回复 | 直到 6 年前

Nkosi 6 年前

所需的JSON错误响应模型与使用的类不匹配。

考虑重构类模型以服务于单个职责

public class ResponseData {
    public List<AccountDTO> AccountList { set; get; }
    public string ReturnMessage { set; get; }
}

还可以重构API操作,以根据操作流返回适当的响应。

[HttpGet]
public IHttpActionResult getAllAccounts(string BRN, string CompanyName) {
    try {
        ResponseData response = new ResponseData();
        List<AccountDTO> accountlist = null;
        IAccountInterface usecase = new AccountDAO();
        if (BRN != null && CompanyName != null) {
            accountlist = usecase.getAccountbyBRNCompanyName().Where(p => p.BRN.ToLower().Equals(BRN.ToLower()) && p.CompanyName.ToLower().Contains(CompanyName.ToLower())).ToList();
            response.ReturnMessage = "Data filter by BRN and CompanyName.";
        } else {
            response.ReturnMessage = "Data filter cannot be null.";
        }
        response.AccountList = accountlist;
        return Ok(response);
    }
    catch (Exception e) {
        Console.WriteLine("Error: " + e.Message + ", Stack: " + e.StackTrace);
        var response = new {
            error = new {
                code = "sample code",
                message = e.Message,
                innererror = new {
                    message = e.Message,
                    type = e.GetType().Name,
                    stacktrace = e.StackTrace
                }
            }
        }
        var responseMessage = Request.CreateResponse(HttpStatusCode.InternalServerError, response);
        return ResponseMessage(responseMessage);
    }
}

NewfrontSolutions 6 年前

        Console.WriteLine("Error: " + e.Message + ", Stack: " + e.StackTrace);
        response.error.code = "sample code";
        response.error.message = e.Message;
        response.error.innererror = "{";
        response.error.message = e.Message;
        response.error.type = "ExceptionType + ...";
        response.error.stacktrace = e.StackTrace;