设计一个预先弹出的屏幕 [英] designing an advance pop up scrren

问题描述

当任何访客首次打开我的网站时，我想弹出一个欢迎屏幕.

我在网站上有四个页面(主页，教程，活动，书籍).访客首次访问任何页面时，都应该获得欢迎"弹出窗口.访问任何页面后，导航到其他页面时都不会再次出现欢迎屏幕.

再次，当访问者重新访问该网站时，应重复该过程.

现在我的查询是:
1.如何建立这样的弹出窗口?

2.我可以使用javascript吗?但是如果禁用了javascript怎么办，还有其他类似ASP.NET的方法吗?

3.我可以选择通过javascript显示弹出窗口，但是如果禁用了javascript，则应该以其他方式显示弹出窗口???

I want to pop up a welcome screen when any visitor opens my website for the first time.

I''ve four pages (home, tutorials, events, books)in my website. The visitor should get the welcome pop up window, when he visits any of the page for the first time. After he visits any of the page, he should not get the welcome screen again when navigating to other pages.

Again when the visitor re-visits the website, the process should repeat.

Now my queries are:
1.How to do build such pop up window?

2.Can I do it with javascript? but what if javascript is disabled, what are the other ways to do this like ASP.NET?

3.Can I make such choice that the pop up window be shown via javascript but if javascript is disabled pop up window should be shown with the other way????

推荐答案

IMO，我不必担心JavaScript会被禁用.每个现代的浏览器都具有它，并且99.9％的用户期望使用它带来高度交互的功能.将没有引导用户的用户重定向到说明如何打开该消息页面以及他们错过的好处的消息页面.

使用会话变量设置一个标志，指示用户是否在第一次访问中，并相应地显示弹出窗口.您可以创建一个显示弹出窗口的JavaScript方法，并根据需要使用RegisterStartupScript对其进行调用.

IMO I wouldn''t worry about JavaScript being disabled. Every modern browser has it and 99.9% of user expect the highly interactive features that come from using it. Redirect the users who don''t to a message page explaining how to turn it on and the benefits they are missing out on.

Use a session variable to set a flag whether the user is on the first visit and display the popup accordingly. You can create a JavaScript method that displays the popup and call it using RegisterStartupScript as necessary.

不是您想要的，而是 ^ ]可能会有所帮助.

单击弹出窗口上的按钮，并确保不会再次调用它.

Not exactly what you want, but This[^] may help.

Click a button on the pop-up and make sure it is not called again.

frnd，

有多种实现此功能的方法.

1)如果访客第一次访问网站，首先您将在访客计算机上创建一个cookie.然后当他下次打开时，您可以读取cookie(如果不存在)，然后将使用j-query或JavaScript显示弹出窗口.

2)您可以跟踪请求的IP地址.您可以获取IP地址并将其存储到数据库中，如果未在数据库中找到请求的IP地址，则可以使用j-Query或Java脚本显示弹出窗口.

对于弹出窗口，您可以使用Gray-box(可以从Google找到)或任何其他JavaScript模型弹出窗口.

让我知道此提示对您是否有帮助.

hi frnd,

there is a multiple way of doing this functionality.

1) first you will create a cookie on the visitor machine if visitor visit website first time . then when he open next time you can read the cookie if not exist then you will show the popup using j-query or JavaScript .

2) you can trace the ip address of the request .you can get the IP address and store into the database and if requested IP address not found into database then you can show the popup using j-Query or java script.

For popup window you can use Gray-box( you can find from the Google) or any other JavaScript model popup windows.

Let Me know if this tip will helpful to you.

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