为什么需要Json请求行为?
如果我想限制HttpGet请求到我的动作，我可以用[HttpPost]属性装饰这个动作
例子:
[HttpPost]
public JsonResult Foo()
{
return Json("Secrets");
}
// Instead of:
public JsonResult Foo()
{
return Json("Secrets", JsonRequestBehavior.AllowGet);
}
为什么[HttpPost]不是充分的? 为什么框架用JsonRequestBehavior“bug”我们。AllowGet为每个JsonResult。如果我想拒绝get请求，我会添加HttpPost属性。
当前回答
你不需要它。
如果你的动作有HttpPost属性，那么你不需要费心设置JsonRequestBehavior并使用没有它的重载。没有JsonRequestBehavior枚举的每个方法都有一个重载。他们是:
没有JsonRequestBehavior
protected internal JsonResult Json(object data);
protected internal JsonResult Json(object data, string contentType);
protected internal virtual JsonResult Json(object data, string contentType, Encoding contentEncoding);
与JsonRequestBehavior
protected internal JsonResult Json(object data, JsonRequestBehavior behavior);
protected internal JsonResult Json(object data, string contentType,
JsonRequestBehavior behavior);
protected internal virtual JsonResult Json(object data, string contentType,
Encoding contentEncoding, JsonRequestBehavior behavior);
其他回答
你不需要它。
如果你的动作有HttpPost属性，那么你不需要费心设置JsonRequestBehavior并使用没有它的重载。没有JsonRequestBehavior枚举的每个方法都有一个重载。他们是:
没有JsonRequestBehavior
protected internal JsonResult Json(object data);
protected internal JsonResult Json(object data, string contentType);
protected internal virtual JsonResult Json(object data, string contentType, Encoding contentEncoding);
与JsonRequestBehavior
protected internal JsonResult Json(object data, JsonRequestBehavior behavior);
protected internal JsonResult Json(object data, string contentType,
JsonRequestBehavior behavior);
protected internal virtual JsonResult Json(object data, string contentType,
Encoding contentEncoding, JsonRequestBehavior behavior);
默认Jsonresult“Deny get”
假设我们有如下方法
[HttpPost]
public JsonResult amc(){}
默认为“Deny Get”。
在下面的方法中
public JsonResult amc(){}
当你需要allowget或使用get时，我们必须使用JsonRequestBehavior.AllowGet。
public JsonResult amc()
{
return Json(new Modle.JsonResponseData { Status = flag, Message = msg, Html = html }, JsonRequestBehavior.AllowGet);
}
MVC默认使用DenyGet来保护您免受涉及JSON请求的非常特定的攻击，以提高在允许HTTP GET暴露之前考虑其影响的可能性。
与之相对的是事后，那时可能已经太迟了。
注意:如果您的action方法不返回敏感数据，那么允许get应该是安全的。
进一步阅读我的Wrox ASP。NET MVC3书
By default, the ASP.NET MVC framework does not allow you to respond to an HTTP GET request with a JSON payload. If you need to send JSON in response to a GET, you'll need to explicitly allow the behavior by using JsonRequestBehavior.AllowGet as the second parameter to the Json method. However, there is a chance a malicious user can gain access to the JSON payload through a process known as JSON Hijacking. You do not want to return sensitive information using JSON in a GET request. For more details, see Phil's post at http://haacked.com/archive/2009/06/24/json-hijacking.aspx/ or this SO post. Haack, Phil (2011). Professional ASP.NET MVC 3 (Wrox Programmer to Programmer) (Kindle Locations 6014-6020). Wrox. Kindle Edition.
StackOverflow相关问题
对于大多数最新的浏览器(从Firefox 21、Chrome 27或IE 10开始)，这不再是一个漏洞。
改进了@Arjen de Mooij的回答 AllowJsonGetAttribute适用于mvc-controllers(不仅仅是单个动作方法):
using System.Web.Mvc;
public sealed class AllowJsonGetAttribute : ActionFilterAttribute, IActionFilter
{
void IActionFilter.OnActionExecuted(ActionExecutedContext context)
{
var jsonResult = context.Result as JsonResult;
if (jsonResult == null) return;
jsonResult.JsonRequestBehavior = JsonRequestBehavior.AllowGet;
}
public override void OnResultExecuting(ResultExecutingContext filterContext)
{
var jsonResult = filterContext.Result as JsonResult;
if (jsonResult == null) return;
jsonResult.JsonRequestBehavior = JsonRequestBehavior.AllowGet;
base.OnResultExecuting(filterContext);
}
}
为了让你自己更容易，你也可以创建一个actionfilterattribute
public class AllowJsonGetAttribute : ActionFilterAttribute
{
public override void OnResultExecuting(ResultExecutingContext filterContext)
{
var jsonResult = filterContext.Result as JsonResult;
if (jsonResult == null)
throw new ArgumentException("Action does not return a JsonResult,
attribute AllowJsonGet is not allowed");
jsonResult.JsonRequestBehavior = JsonRequestBehavior.AllowGet;
base.OnResultExecuting(filterContext);
}
}
把它用在你的行动上
[AllowJsonGet]
public JsonResult MyAjaxAction()
{
return Json("this is my test");
}
