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我希望你能帮助我解决一个难题。在我在下面问我的问题之前，我要写一堆代码:)

在我的页面上单击某些内容时：

<input type='button' value='1' name='derp1' onclick='OpenTelerikWindow(1)' />
<br/>
<input type='button' value='2' name='derp2' onclick='OpenTelerikWindow(2)' />

我用 Jquery 打开 Telerik 窗口：

function OpenTelerikWindow(arg) {
var url = '/Controller/Derp/';
$.ajax({
    type: "GET",
    url: url,
    dataType: "html",
    data: { id: arg }
    success: function (data) {
        $('#PaymentWindow').data("tWindow").content(data);
        $('#PaymentWindow').data("tWindow").center().open().refresh();
    }
  });
}

我的控制器 ActionResult：

public ActionResult Derp(int id)
{
    SomeModel someModel = _GetModel(id);
    return PartialView("Derp", someModel)
}

然后我的 Telerik Window 的内容是这样的：

@SomeModel
<div id="theDerpina">
    <div>
        //Some Stuff
        @using (Ajax.BeginForm("Derpina1", "Controller", new { id = SomeModel.id }, new AjaxOptions
            {
                HttpMethod = "POST",
                UpdateTargetId = "theDerpina",
                InsertionMode = InsertionMode.Replace
            }, new { id = "feedback-form" }))
        {
            //Some Stuff
            <button type="submit" > Submit</button>
            <br/>
            <button type="button" > CloseWindow </button>
        }
    </div>
    <div>
        //More Stuff
        @using (Ajax.BeginForm("Derpina2", "Controller", new { id = SomeModel.id }, new AjaxOptions
            {
                HttpMethod = "POST",
                UpdateTargetId = "theDerpina",
                InsertionMode = InsertionMode.Replace
            }, new { id = "feedback-form" }))
        {
            //Different Stuff
            <button type="submit" > Submit</button>
            <br/>
            <button type="button" > CloseWindow </button>
        }
    </div>
</div>

我的另外两个控制器动作：

public ActionResult Derpina1(int id)
{
    SomeModel someModel = _GetModel(id);
    if(ModelState.IsValid)
    {
        //DoStuff
        return View("SomeOtherView");
    }
    else
    {
        return PartialView("Derp", someModel);
    }
}

public ActionResult Derpina2(int id)
{
    SomeModel someModel = _GetModel(id);
    if(ModelState.IsValid)
    {
        //DoDifferentStuff
        return View("SomeOtherView");
    }
    else
    {
        return PartialView("Derp", someModel);
    }
}

当我打开一次窗口时，一切正常。但是，如果我要打开窗口，关闭它，然后再次打开它，就会发生奇怪的事情。例如，假设我单击了 Derpina1 的提交按钮，我将收到两个对该特定控制器操作的调用。如果我在 Firebug 中监控控制台，我会看到对控制器操作的两个单独调用。如果我再次关闭窗口，再次打开它，然后再次提交，同样的事情会发生，我的控制器操作现在将收到 4 次对我的控制器操作的调用。

也许你们可以指出正确的方向。我是否应该以不同的方式打开 Telerik 窗口，如果发生这种情况，我是否应该使用不同的方法返回 ModelError？（因为如果我收到 ModelError，我将面临同样的行为）

4

1 回答 1

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I managed to find out the solution after googling a while. All I did was creating a Window on the fly from within the script, and then destroying it after I have used it, like so:

function OpenTelerikWindow(arg) {
var url = '/Controller/Derp/';
$.ajax({
    type: "GET",
    url: url,
    dataType: "html",
    data: { id: arg }
    success: function (data) {
        var paymentWindow = $("<div id='PaymentWindow'></div>").tWindow({
            title: "Payment",
            contentUrl: '',
            html: data,
            modal: true,
            resizable: false,
            draggable: true,
            width: 500,
            height: 640,
            onClose: function (e) {
                e.preventDefault();
                paymentWindow.data('tWindow').destroy();
            }
        });
        paymentWindow.data('tWindow').center().open();
    }
  });
}
于 2012-11-20T15:54:34.280 回答