Question

首先，抱歉我的英语不好。

我希望当一个人点击“Page1”按钮时，控制器会返回“Page1”的渲染部分以及“page2”和“allPage”的相同内容。

我的观点是：

@{
ViewBag.Title = "Title";
}
<script src="@Url.Content("~/Scripts/jquery.unobtrusive-ajax.min.js")" type="text/javascript"></script>
<h2>Title</h2>

@using (Ajax.BeginForm("NameAction", //controller action name
"NameController", //controller name
new AjaxOptions //ajax options that tell mvc how to perform the replacement
{
UpdateTargetId = "ViewPage", //id of div to update
HttpMethod = "Post" //how to call the controller action
}, new { id = "FormName" }))
{
<input type="submit" id="btn" value="p1" id="p1"/>
<input type="submit" id="btn" value="p2" id="p2"/>
<input type="submit" id="btn" value="AllPage" id="AllPage"/>
<div id="ViewPage">
    //render partial view
</div>        
}

我的控制器是：

public class NameController : Controller
{
    [HttpPost]
    public ActionResult NameAction(String btn)
    {
        if (Request.IsAjaxRequest())
        if(btn="p1")
        return PartialView("p1");
        if(btn="2")
        return PartialView("p2");
        if(btn="3")
        return PartialView("p3");


        return View();
    }
}

Request.isAjaxRequest等于总是false，而partialview不更新div但是擦除所有页面

感谢您的帮助。

Answer 1

为您的提交按钮命名：

<input type="submit" name="btn" value="page1" id="Page1"/>
<input type="submit" name="btn" value="Page2" id="Page2"/>
<input type="submit" name="btn" value="AllPage" id="AllPage"/>

然后：

[HttpPost]
public ActionResult NameAction(string btn)
{
    if (btn == "page1")   
    {
        // the page1 button was clicked
    } 
    else if (btn == "page2")
    {
        // the page2 button was clicked
    }
    else if (btn == "AllPage")
    {
        // the AllPage button was clicked
    }

    ...
}

如果您不想依赖按钮的实际标签：

<button type="submit" name="btn" value="p1" id="Page1">Show page 1</button>
<button type="submit" name="btn" value="p2" id="Page2">Show page 2</button>
<button type="submit" name="btn" value="all" id="AllPage">Show all pages</button>

在控制器中你可以检查值。

更新：

确保已将jquery.unobtrusive-ajax脚本包含在页面中，以便Ajax.BeginForm可以工作并发送AJAX请求：

<script src="@Url.Content("~/Scripts/jquery.unobtrusive-ajax.min.js")" type="text/javascript"></script>

如何使用ajax.BeginForm更改内容？

1 个答案: