本文介绍了从下拉列表中选择项目时,Ajax弹出窗口关闭的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
Div作为弹出窗口.问题是:
我使用了一个div作为弹出窗口,其中包含一些texbox,下拉列表和按钮.
每次在下拉列表中选择一个项目时,div就会消失.
如何阻止div关闭?
每次div消失时都会选择下拉项,这就是为什么我再次打开div的原因
字符串脚本= " < script language ='javascript'> fnOpenDiv();</script>"; ScriptManager.RegisterStartupScript(Page,Page.GetType(),Guid.NewGuid().ToString(),script, false );
Div as a popup. problem is :
I have a used a div as pop up that contains some texboxes, dropdownlists and button.
each time an item in a dropdown list is selected a div dissapears.
how to stop the div from being closed?
The dropdown item is get selected each time the div dissapears that''s why i opened the div again by
string script = "<script language='javascript'>fnOpenDiv();</script>"; ScriptManager.RegisterStartupScript(Page, Page.GetType(), Guid.NewGuid().ToString(), script, false);
解决方案
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08-13 13:53