Ajax.BeginForm 提交后产生两个表单

Question

Model 类

C# code

public class Department {
    public string Name { get; set; }
    public string Description { get; set; }
}

Controller 类

C# code

public class DepartmentController : Controller
{
    [HttpGet]
    public ActionResult Create() {
        return View();
    }
    
    [HttpPost]
    public ActionResult Create(TestEntity.Department department) {
        ViewBag.MyMessage = "Error Message1";
        return View();
    }
}

View 

View

@model TestEntity.Department
@{
    ViewBag.Title = "Create Department";
    Layout = null;
}
<script src="@Url.Content("~/Scripts/jquery-1.5.1.min.js")" type="text/javascript"></script>
<script src="@Url.Content("~/Scripts/modernizr-1.7.min.js")" type="text/javascript"></script>
<script src="@Url.Content("~/Scripts/jquery.validate.js")" type="text/javascript"></script> 
<script src="@Url.Content("~/Scripts/jquery.validate.unobtrusive.js")" type="text/javascript"></script> 
<script src="@Url.Content("~/Scripts/jquery.unobtrusive-ajax.js")" type="text/javascript"></script> 
<h2>Create Department</h2>

@using (Ajax.BeginForm("Create", "Department", 
        new AjaxOptions { UpdateTargetId = "divForm", HttpMethod="POST"
        })) {
    <fieldset>
        <legend>Create Department</legend>
    @Html.LabelFor(m => m.Name)
    @Html.TextBoxFor(m => m.Name, new { id="txtName", style="background-color:#CFDFFC"})
    @ViewBag.MyMessage <br />
    @Html.LabelFor(m => m.Description)
    @Html.TextBoxFor(m => m.Description) <br />
    <input type="submit" id="btnSubmit" value="Create" />
    </fieldset>
}
<div id="divForm"></div>

进行异步提交之后，在 <div id="divForm"></div> 内部显示了正确的信息，但是不合理的是，原来的表单还在。如此一来，异步提交之后，页面上显示两个创建部门的表单了

Answer 1

我无法重现你的错误，没有出现两个表单。我的运行情况如下：

如果你的意思是点击Submit之后不需要出现Create页面，那么应该设法使用跳转命令到其它View, 譬如:

 [HttpPost]
        public ActionResult Create(Department department)
        {
            ViewBag.MyMessage = "Error Message1";
            return View("Successful");  //Create Successful page to jump to
        }

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Answer 2

用ajax异步提交的话 你需要在POST的方法里加个判断。

if(!Request.IsAjaxRequest())
	return View();
else
	//你可以返回json也可以输出text或html，随便你做个处理

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