问题描述
我想请求Ajax的弹出BIJ点击与变量的链接。弹出窗口会显示从服务器请求某个日期。提交后的数据将被在服务器上检查，并发送回的反响。根据不同的性反应，新的内容将显示在弹出。用户可以随后关闭弹出窗口。
I like to request a ajax-popup bij clicking on a link with variable. The popup will show some date requested from the server. After submit the data will be checked on the server and send a responce back. Depending on the responce, new content will be showed in the popup. The user can close the popup afterwards.
我已经寻找一个教程或示例一段时间，但我找不到任何。
I have searched a while for a tutorial or example, but i can't find any.
谁能给我一个例子或建议，以启动？
Who can give me an example or suggestion to startup?
推荐答案
请进入 jQuery UI的对话框部件。它可以做你的需要。你只要把这些控件，文字等为一个DIV，然后再调用弹出对话框的方法。由于code是在一个div，就可以动态地为需要AJAX调用的结果进行更新。
Check into the jQuery UI Dialog widget. It can do what you need. You just put the controls, text, etc into a DIV, and then invoke the method to pop the dialog. Since the code is in a div, you can update it dynamically with the results of ajax calls as needed.
这篇关于HOWTO创建一个Ajax弹出窗体并提交表单的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！