HOWTO创建一个Ajax弹出窗体并提交表单 [英] howto create an Ajax Popup Form and submit the form

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.

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