将PartialView显示为弹出窗口

问题描述

我认为我有

@foreach (var item in Model)
{
    <li>@Html.ActionLink("click me", "popup", "SomeData", new{id = item.ID}, new {@class = "PopUp"})</li>
}

然后我有一个看起来像这样的控制器；

I then have a controller that looks like this;

public ActionResult popup(Guid id)
{
    var singelData = db.SomeRandomData.Find(id);

    return PartialView(singelData);
}

还有一个看起来像这样的partialView；

And a partialView that looks like this;

<div>This a popup</div>
@Model.metadata1

到目前为止，当我单击链接时，我被重定向到partialView.

So far so good, when I click on a link I get redirected to a partialView.

现在我不舒服的部分是脚本部分，这是我的尝试；

Now to the part where I'm not comfortable, the script section, here is my attempt;

<script src="~/Scripts/jquery-ui-1.8.20.js"></script>
<script type="text/javascript">
    $(function () {
        $('.PopUp').click(function () {
            $('<div/>').appendTo('body').dialog({
                close: function (event, ui) {
                    dialog.remove();
                },
                modal: true
            }).load(this.href, {});

            return false;
        });
    });
</script>

但是它仍然只返回一个视图.我哪里出问题了?

But it still just returns a view. Where do i go wrong?

推荐答案

您应确保在显示的2个脚本之前 包含jQuery.另外，我建议您使用诸如FireBug之类的JavaScript调试工具，并查看控制台窗口，在该窗口中可能会看到脚本潜在的错误.

You should make sure that you have jQuery included before the 2 scripts you have shown. Also I would recommend you using a javascript debugging tool such as FireBug and looking at the console window where you might see potential errors with your scripts.

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