In ASP.NET MVC beta, we can use Ajax.BeginForm, Ajax.ActionLink to make Ajax calls. Similarly, we can also use some Ajax-supported frameworks such as jQuery to simplify ajax calls.
1. Using System.Web.Mvc.Ajax
1.1 System.Web.Mvc.Ajax.BeginForm
1.2 System.Web.Mvc .Ajax.ActionLink
2. Manually create your own "non-intrusive" Javascript"
1. Use System.Web.Mvc.Ajax

1.1 System.Web.Mvc.Ajax.BeginForm
Step 1: Create Form## with Ajax.BeginForm
#
@using (Ajax.BeginForm(
new AjaxOptions()
{
HttpMethod = "post",
Url = @Url.Action("Index","Reviews"),
InsertionMode = InsertionMode.Replace,
UpdateTargetId = "restaurantList",
LoadingElementId = "loding",
LoadingElementDuration = 2000
}))
{
<input type="search" name="searchItem"/>
<input type="submit" value="按名称搜索"/>
}
The final generated form is as follows:
<form id="form0" method="post"
data-ajax-url="/Reviews"
data-ajax-update="#restaurantList"
data-ajax-mode="replace"
data-ajax-method="post"
data-ajax-loading-duration="2000"
data-ajax-loading="#loding"
data-ajax="true"
action="/Reviews" novalidate="novalidate">
Step 2: Create Ajax.BeginForm The new AjaxOptions() object's Url points to Action
new AjaxOptions()
{
...
Url = @Url.Action("Index","Reviews")
...
}
public ActionResult Index(string searchKey = null)
{
var model = _restaurantReviews.Where(r => searchKey == null || r.Name.ToLower().Contains(searchKey.ToLower().Trim()))
.OrderByDescending(r => r.Rating)
.Take(100)
.Select(r=>new RestaurantReview()
{
City = r.City,
Country = r.Country,
Id = r.Id,
Name = r.Name,
Rating = r.Rating
}).ToList();
if (Request.IsAjaxRequest())
{
System.Threading.Thread.Sleep(1000 * 3);//模拟处理数据需要的时间
//return View(model)会返回整个页面,所以返回部分视图。
return PartialView("_RestaurantPatialView", model);
}
return View(model);
}
Note: Instructions on using System.Web.Mvc.Ajax: Action method of Controller: (1) When [HttpPost] is explicitly added, the HttpMethod passed to AjaxOptions() of System.Web.Mvc.Ajax can only be "post", himgate in us in in in\\Ajax[HttpGet], The HttpMethod passed to AjaxOptions() of System.Web.Mvc.Ajax can only be "get", (3) When all Without explicitly adding [HttpPost] and [HttpGet], the HttpMethod passed to AjaxOptions() of System.Web.Mvc.Ajax can be "get" or "post", :
<p id="restaurantList">... </p>
## Step 4: (Optional) To enhance the user experience, add an AjaxOption object The ID specified by the LoadingElementid parameter is the HTML element of Loding:
CSHTML files: 
r 动
#1.2 System.Web.Mvc.Ajax.ActionLink
The usage of System.Web.Mvc.Ajax.ActionLink and System.Web.Mvc.Ajax.BeginForm are basically the same
The first step: Use System.Web.Mvc.Ajax.ActionLink to create a hyperlink
new AjaxOptions()
{
....
LoadingElementId = "loding",
LoadingElementDuration = 2000
}))
The corresponding generated final html is:
<p id="loding" hidden="hidden">
<img class="smallLoadingImg lazy" src="/static/imghwm/default1.png" data-src="@Url.Content(" ~/Content/images/loading.gif")" / alt="A brief analysis of the use of Ajax in Asp.net MVC" >
</p>
Step 2: Define the Action to respond to the hyperlink:
@*@Html.ActionLink(item.Name, "Details", "Reviews",new{id = item.Id},new {@class ="isStar"})*@
@*<a class="isStar" href="@Url.Action("Details","Reviews", new {id = item.Id})">@item.Name</a>*@
@*使用Ajax的超链接*@
@{
var ajaxOptions = new AjaxOptions()
{
HttpMethod = "post",
//Url = @Url.Action(""),
UpdateTargetId = "renderBody",
InsertionMode = InsertionMode.Replace,
LoadingElementId = "loding",
LoadingElementDuration = 2000
};
@Ajax.ActionLink(item.Name, "Details", "Reviews", new { id = item.Id }, ajaxOptions, new {@class="isStar"})
}
Step 3: Define the bearer Update some html elements:
<a class="isStar" href="/Reviews/Details/1" data-ajax-update="#renderBody" data-ajax-mode="replace" data-ajax-method="post" data-ajax-loading-duration="2000" data-ajax-loading="#loding" data-ajax="true">
## Step 4: (Optional) To enhance the user experience, add the AjaxOptionsd object The Id specified by the LoadingElementId parameter is the loading html element:
/// <summary>
///关于使用System.Web.Mvc.Ajax的说明:
/// Controller的Action方法:
/// (1)当显式添加[HttpPost],传给System.Web.Mvc.Ajax的AjaxOptions()的HttpMethod只能为 "post",
/// (2)当显式添加[HttpGet],传给System.Web.Mvc.Ajax的AjaxOptions()的HttpMethod只能为 "get",
/// (3) 当都没有显式添加[HttpPost]和[HttpGet],传给System.Web.Mvc.Ajax的AjaxOptions()的HttpMethod可以为 "get"也可以为"post",
/// </summary>
/// <param name="id"></param>
/// <returns></returns>
public ActionResult Details(int id=1)
{
var model = (from r in _restaurantReviews
where r.Id == id
select r).FirstOrDefault();
if (Request.IsAjaxRequest())
{
return PartialView("_RestaurantDetails", model);
}
return View(model);
}
The generated form is:
<p id="renderBody">
....
</p>
Step 2: Add Action to process the form: This is the same as the second step in 1.1. Step 3: Add Js to process the form:
@* ---------------------------------------------------------
需要手工为Form添加些属性标签,用于锚点
模仿MVC框架的构建自己的“非介入式Javascript”模式
-------------------------------------------------------*@
<form method="post"
action="@Url.Action("Index")"
data-otf-ajax="true"
data-otf-ajax-updatetarget="#restaurantList">
<input type="search" name="searchItem" />
<input type="submit" value="按名称搜索" />
</form>
The above is what I compiled for everyone. I hope it will be helpful to everyone in the future. Related articles:
JSONP for AJAX cross-domain request to obtain JSON data
Realize mobile phone positioning based on h5 ajax
Use the H5 feature FormData to upload files without refreshing
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