如何让 ASPxPopupControl 从(以前)隐藏按钮中出现?
我想在用户单击按钮时显示 ASPxPopupControl。但这有一个问题。该按钮属于一个面板,默认情况下是隐藏的。仅当用户告诉它这样做时才会显示。
如果面板最初可见,则弹出控件显示得很好。但如果它最初是不可见的(由属性 Visible="false"
设置),那么即使面板显示,弹出控件也永远不会出现!
我该如何解决这个问题?我知道我应该在 Devexpress 论坛上问这个问题,但如果我在这里尝试也没什么坏处,对吧?
谢谢你!
I want to show up an ASPxPopupControl whenever an user click a button. But there's a problem with this. The button belongs to a panel, which is hidden by default. It's only shown when user tells it to do so.
If the panel is initially visible, the popup control shows just fine. But if it's initially invisible (set by attribute Visible="false"
), then when even the panel shows up, the popup control never appears!
How can I fix this? I know that I should ask this question in Devexpress forums, but it never hurts if I try here, right?
Thank you!
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您使用的是
ASPxPanel
控件吗?如果是这样,请尝试使用面板的ClientVisible
属性而不是Visible
。这有帮助吗?更新
我建议您开始使用
ASPxPanel
并使用其ClientVisible
属性。如果您想使用标准Panel
,则应使用style.visibility
属性隐藏它:C#:
Are you using the
ASPxPanel
control? If so, try to use the panel'sClientVisible
property instead ofVisible
. Does this help?UPDATE
I would suggest that you start using the
ASPxPanel
instead and use itsClientVisible
property. If you want to use the standardPanel
, you should hide it using thestyle.visibility
property:C#:
我不确定,但我认为通过在服务器端代码中放置 Visible = false 不会将控件呈现给浏览器。所以我们无法通过 DOM 访问它。我认为你需要设置样式而不是 Visible 属性。
I am not sure, but I think by putting Visible = false in the server side code will not render the control to the browser. So we can't access it via DOM. I think you need to set the style instead of Visible property.