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);
}

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;


        }

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);
}

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())

在action方法中，返回Json(object)将Json返回到页面。

public ActionResult SomeActionMethod() {
  return Json(new {foo="bar", baz="Blech"});
}

然后使用Ajax调用action方法。您可以使用ViewPage中的一个辅助方法，例如

<%= Ajax.ActionLink("SomeActionMethod", new AjaxOptions {OnSuccess="somemethod"}) %>

SomeMethod将是一个javascript方法，然后计算返回的Json对象。

如果你想返回一个普通的字符串，你可以使用ContentResult:

public ActionResult SomeActionMethod() {
    return Content("hello world!");
}

ContentResult默认返回一个文本/纯文本作为它的contentType。 这是可重载的，所以你还可以做:

return Content("<xml>This is poorly formatted xml.</xml>", "text/xml");

另一种处理JSON数据的好方法是使用JQuery getJSON函数。你可以致电

public ActionResult SomeActionMethod(int id) 
{ 
    return Json(new {foo="bar", baz="Blech"});
}

方法从jquery getJSON方法简单…

$.getJSON("../SomeActionMethod", { id: someId },
    function(data) {
        alert(data.foo);
        alert(data.baz);
    }
);