与编码框架的替代解决方案

操作返回json

控制器

    [HttpGet]
    public ActionResult SomeActionMethod()
    {
        return IncJson(new SomeVm(){Id = 1,Name ="Inc"});
    }

剃须刀页面

@using (var template = Html.Incoding().ScriptTemplate<SomeVm>("tmplId"))
{
    using (var each = template.ForEach())
    {
        <span> Id: @each.For(r=>r.Id) Name: @each.For(r=>r.Name)</span>
    }
}

@(Html.When(JqueryBind.InitIncoding)
  .Do()
  .AjaxGet(Url.Action("SomeActionMethod","SomeContoller"))
  .OnSuccess(dsl => dsl.Self().Core()
                              .Insert
                              .WithTemplate(Selector.Jquery.Id("tmplId"))
                              .Html())
  .AsHtmlAttributes()
  .ToDiv())

操作返回html

控制器

    [HttpGet]
    public ActionResult SomeActionMethod()
    {
        return IncView();
    }

剃须刀页面

@(Html.When(JqueryBind.InitIncoding)
  .Do()
  .AjaxGet(Url.Action("SomeActionMethod","SomeContoller"))
  .OnSuccess(dsl => dsl.Self().Core().Insert.Html())
  .AsHtmlAttributes()
  .ToDiv())

我发现用JQuery实现MVC ajax GET调用的几个问题，这让我头疼，所以在这里分享解决方案。

Make sure to include the data type "json" in the ajax call. This will automatically parse the returned JSON object for you (given the server returns valid json). Include the JsonRequestBehavior.AllowGet; without this MVC was returning a HTTP 500 error (with dataType: json specified on the client). Add cache: false to the $.ajax call, otherwise you will ultimately get HTTP 304 responses (instead of HTTP 200 responses) and the server will not process your request. Finally, the json is case sensitive, so the casing of the elements needs to match on the server side and client side.

JQuery示例:

$.ajax({
  type: 'get',
  dataType: 'json',
  cache: false,
  url: '/MyController/MyMethod',
  data: { keyid: 1, newval: 10 },
  success: function (response, textStatus, jqXHR) {
    alert(parseInt(response.oldval) + ' changed to ' + newval);                                    
  },
  error: function(jqXHR, textStatus, errorThrown) {
    alert('Error - ' + errorThrown);
  }
});

示例MVC代码:

[HttpGet]
public ActionResult MyMethod(int keyid, int newval)
{
  var oldval = 0;

  using (var db = new MyContext())
  {
    var dbRecord = db.MyTable.Where(t => t.keyid == keyid).FirstOrDefault();

    if (dbRecord != null)
    {
      oldval = dbRecord.TheValue;
      dbRecord.TheValue = newval;
      db.SaveChanges();
    }
  }

    return Json(new { success = true, oldval = oldval},
                JsonRequestBehavior.AllowGet);
}

我认为您应该考虑请求的AcceptTypes。我在我当前的项目中使用它来返回正确的内容类型，如下所示。

你在控制器上的动作可以像在请求对象上一样测试它

if (Request.AcceptTypes.Contains("text/html")) {
   return View();
}
else if (Request.AcceptTypes.Contains("application/json"))
{
   return Json( new { id=1, value="new" } );
}
else if (Request.AcceptTypes.Contains("application/xml") || 
         Request.AcceptTypes.Contains("text/xml"))
{
   //
}

然后，您可以实现视图的aspx来满足部分xhtml响应的情况。

然后在jQuery中，你可以将类型参数作为json来获取它:

$.get(url, null, function(data, textStatus) {
        console.log('got %o with status %s', data, textStatus);
        }, "json"); // or xml, html, script, json, jsonp or text

你可能想看看这篇非常有用的文章，它很好地介绍了这一点!

只是觉得它可能会帮助人们找到解决这个问题的好办法。

http://weblogs.asp.net/rashid/archive/2009/04/15/adaptive-rendering-in-asp-net-mvc.aspx

PartialViewResult和jsonresult继承自基类ActionResult。因此，如果返回类型是动态决定的，则将方法输出声明为ActionResult。

public ActionResult DynamicReturnType(string parameter)
        {
            if (parameter == "JSON")
                return Json("<JSON>", JsonRequestBehavior.AllowGet);
            else if (parameter == "PartialView")
                return PartialView("<ViewName>");
            else
                return null;


        }

    public ActionResult GetExcelColumn()
    {            
            List<string> lstAppendColumn = new List<string>();
            lstAppendColumn.Add("First");
            lstAppendColumn.Add("Second");
            lstAppendColumn.Add("Third");
  return Json(new { lstAppendColumn = lstAppendColumn,  Status = "Success" }, JsonRequestBehavior.AllowGet);
            }
        }