从服务器回叫到客户端 [英] Call back from server to client

问题描述

我正在使用ASP.NET MVC 4应用程序，我需要通过从Controller向客户端发送消息来在客户端中显示消息.

I am using ASP.NET MVC 4 application, I need to Display messages in the Client, by sending messages from Controller to Client.

我的要求是用户单击UI中的按钮，然后我将处理服务器上的文件，并在我处理的每个foreach文件的末尾在UI中显示消息.我需要在使用ASP.NET MVC的客户端中显示文件名.

My requirement is user click a button in UI and i will process the files on the server and Display message in UI on end of each foreach file i process. i need to show the File names in the Client Using ASP.NET MVC.

任何人都可以通过每次在for-each循环上从服务器调用客户端方法来帮助如何在客户端中显示消息.

Can any one Help how to show the messages in the Client by calling client method from server on for-each loop each time.

我能够调用控制器并将每个控制器的末尾发送最终消息到UI，但是如何在每个foreach循环迭代中发送?

I am able to call the controller and end of each controller I am sending final message to UI, but how to send on each foreach loop iteration?

推荐答案

您必须编写一个ActionResult，以便逐步将结果写入响应中.因此您可以在每个foreach循环迭代中向用户显示一些数据.我写了一个简单的ActionResult，每2秒写一个数字:

You have to write an ActionResult that progressively write result to the response. so you can show the user some data in every foreach loop iteration. I have written a simple ActionResult that writes a number every 2 seconds:

public class ProgressiveResult : ActionResult
{
    public override void ExecuteResult(ControllerContext context)
    {
        for (int i = 0; i < 20; i++)
        {
            context.HttpContext.Response.Write(i.ToString());
            Thread.Sleep(2000);
            context.HttpContext.Response.Flush();
        }
        context.HttpContext.Response.End();
    }
}

这是一个返回此结果的动作:

and this is an action that returns this result:

public ActionResult LongProcess()
{
   return new ProgressiveResult();
}

因此您可以编写ActionResult并使用ExecuteResult方法编写foreach代码.

So you can write an ActionResult and write your foreach code in ExecuteResult method.

更新:

您可以使用Ajax请求进行此调用，并使用简单的代码(如以下代码)返回结果:

You can make this call with an Ajax request and return result with a simple code like the following code:

var result = "";
function showResult() {
    if (result !== oReq.responseText) {
        result = oReq.responseText;
        console.log(result);
    }
}
var oReq = new XMLHttpRequest();
oReq.open("get", "/Home/LongProcess", true);
oReq.send();
setInterval(showResult, 1000);

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