如何使用动作过滤器在 asp.net mvc 中集中模型状态验证?

问题描述

我在几个地方写了这段代码并且总是重复这个逻辑:

I write this code in several places and always repeat this logic:

public ActionResult MyMethod(MyModel collection)
{
    if (!ModelState.IsValid)
    {
        return Json(false);//to read it from javascript, it's always equal
    }
    else
    {
        try
        {
            //logic here
            return Json(true);//or Json(false);
        }
        catch
        {
            return Json(false);//to read it from javascript, it's always equal
        }
    }
}

有什么方法可以使用动作过滤器，不要重复try-catch，询问模型是否有效，返回Json(false)为ActionResult?

Is there any way using action filters, not to be repeating the try-catch, ask if the model is valid and return Json(false) as ActionResult?

推荐答案

为了符合 REST，你应该返回 http bad request 400 表示请求格式错误(模型无效)而不是返回 Json(false).

To conform with REST, you should return http bad request 400 to indicate that the request is malformed (model is invalid) instead of returning Json(false).

从 asp.net 官方网站 用于 web api:

Try this attribute from asp.net official site for web api:

public class ValidateModelAttribute : ActionFilterAttribute
{
     public override void OnActionExecuting(HttpActionContext actionContext)
     {
        if (!actionContext.ModelState.IsValid)
        {
            actionContext.Response = actionContext.Request.CreateErrorResponse(
                HttpStatusCode.BadRequest, actionContext.ModelState);
        }
    }
}

asp.net mvc 的版本可能是这样的:

A version for asp.net mvc could be like this:

public class ValidateModelAttribute : ActionFilterAttribute
{
        public override void OnActionExecuting(ActionExecutingContext filterContext)
        {
              if (!filterContext.Controller.ViewData.ModelState.IsValid)
              {
                   filterContext.Result = new HttpStatusCodeResult(HttpStatusCode.BadRequest);
              }
        }
}

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