AJAX弹出窗口的asp.net [英] AJAX Pop up window for asp.net

问题描述

我正在使用C＃在asp.net上为医院制作新项目。我想显示警报弹出窗口，如果用户在任何页面中警报弹出窗口将在每10分钟后自动显示关于从约会表中预约患者。当用户关闭它然后再次显示10分钟后弹出窗口将显示，当点击提醒稍后按钮时，弹出窗口将不会显示给该用户。当再次登录软件时，每个登录的用户都会重复相同的程序。

解决方案

你可以使用Timer控件。



看看这些：

Javascript计时器 [ ^ ]

Ajax Timer控件 [ ^ ]

Ajax Timer control - 2 [ ^ ]

MSDN：定时器控制概述 [ ^ ]



根据计时器控制，显示警报窗口。根据窗口上的选择，再次设置计时器。以下文章可以帮助打开弹出窗口，然后在关闭后从子窗口向父页面发回一个值：在子窗口关闭时通过AJAX（ASP.NET 2.0）对父页面进行部分更新 - 有条件地 [ ^ ]

正如Sandeep所说，Ajax计时器将是采用的好选择。并在cookie / session中存储任何位值（如shopopup = 1或0）。当用户稍后点击提醒时，该位变为0.在显示弹出窗口之前检查此位。如果它只有1然后显示弹出窗口，则无需显示。当用户退出时，此位需要转1.



谢谢，

Rohit

< blockquote>你好 如何ru ??

I am making new project on hospital in asp.net using C#. I want to show the alert popup if user in any page the alert popup window will show automatically after each 10 minutes about appointment of patients from appointment table. when user close it then again after 10 minutes popup window will show and when click on remind later button then popup window will not show to that user. when again log in to the software again same procedure will be repeated for each and every user who logged in.

解决方案

You can use Timer control for it.

Look at these:
Javascript timers[^]
Ajax Timer control[^]
Ajax Timer control - 2[^]
MSDN: Timer Control Overview[^]

Based on timer control, show your alert window. Based on selection made on the window, set the timer again. Following article can be of help for opening a popup and then sending back a value to parent page from child window after close: Partial update of parent page via AJAX (ASP.NET 2.0) on close of a child window - conditionally[^]

Hi,

As Sandeep told, Ajax timer will be good option to adopt. and Store any bit value (like shopopup=1 or 0) in cookie/session. when User will hit remind later, this bit turns 0. Check this bit before showing popup. If it is 1 only then show popup else no need to show. When user will sign out, then this bit needs to turn 1.

Thanks,
Rohit

Hello How r u??

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