C# ASP.NET MVC：如何创建一个动作过滤器来输出 JSON？
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ASP.NET MVC: How to create an action filter to output JSON?
提问by aleemb
My second day with ASP.NET MVC and my first request for code on SO (yep, taking a short cut).
我在 ASP.NET MVC 的第二天和我对 SO 代码的第一个请求（是的，走捷径）。
I am looking for a way to create a filter that intercepts the current output from an Action and instead outputs JSON (I know of alternate approachesbut this is to help me understand filters). I want to ignore any views associated with the action and just grab the ViewData["Output"], convert it to JSON and send it out the client. Blanks to fill:
我正在寻找一种方法来创建一个过滤器，该过滤器拦截来自 Action 的当前输出并输出 JSON（我知道其他方法，但这是为了帮助我理解过滤器）。我想忽略与操作相关的任何视图，只需获取 ViewData["Output"]，将其转换为 JSON 并将其发送给客户端。填空：
TestController.cs:
TestController.cs:
[JSON]
public ActionResult Index()
{
ViewData["Output"] = "This is my output";
return View();
}
JSONFilter.cs:
JSONFilter.cs:
public override void OnActionExecuting(ActionExecutingContext filterContext)
{
/*
* 1. How to override the View template and set it to null?
* ViewResult { ViewName = "" } does not skip the view (/Test/Index)
*
* 2. Get existing ViewData, convert to JSON and return with appropriate
* custom headers
*/
}
Update: Community answers led to a fuller implementation for a filter for JSON/POX.
更新：社区答案导致了 JSON/POX 过滤器的更完整实现。
采纳答案by tvanfosson
I would suggest that what you really want to do is use the Model rather than arbitrary ViewDataelements and override OnActionExecutedrather than OnActionExecuting. That way you simply replace the result with your JsonResultbefore it gets executed and thus rendered to the browser.
我建议您真正想做的是使用 Model 而不是任意ViewData元素并覆盖OnActionExecuted而不是OnActionExecuting. 这样，您只需在结果JsonResult被执行并呈现给浏览器之前将其替换为您的结果。
public class JSONAttribute : ActionFilterAttribute
{
...
public override void OnActionExecuted( ActionExecutedContext filterContext)
{
var result = new JsonResult();
result.Data = ((ViewResult)filterContext.Result).Model;
filterContext.Result = result;
}
...
}
[JSON]public ActionResult Index()
{
ViewData.Model = "This is my output";
return View();
}
回答by JSC
回答by Troy
You haven't mentioned only returning the JSON conditionally, so if you want the action to return JSON every time, why not use:
您没有提到仅有条件地返回 JSON，因此如果您希望操作每次都返回 JSON，为什么不使用：
public JsonResult Index()
{
var model = new{ foo = "bar" };
return Json(model);
}
