选择单选按钮时显示一个框 [英] show a box when radio button selected

问题描述

我有4个单选按钮.
优秀
好
一般
差

我想要一个功能，当用户选择较差时，然后打开一个对话框进行评论.
喜欢
http://www.eurekaforbes.com/our-service.aspx

如果有人对此有任何了解，请告诉我.


请在检查我之前先检查上方的链接，还有我正在使用php进行的其他操作.
谢谢

I have 4 radio button.
excellent
Good
Average
Poor

i want a function that when user select poor then a dialogue box open for comment.
like
http://www.eurekaforbes.com/our-service.aspx

if any one can have any knowledge about this then pls let me know.


Please check the upper link before ans me and one thing more i am working with php.
thanks

推荐答案

尝试以下链接:

运行中的ModalPopupExtender控件 [ http://gugiaji.wordpress.com/2012/03 /19/create-a-popup-dialog-with-modalpopupextender-ajaxcontroltoolkit/ [

hi try these links:

ModalPopupExtender Control in Action[^]


http://gugiaji.wordpress.com/2012/03/19/create-a-popup-dialog-with-modalpopupextender-ajaxcontroltoolkit/[^]

hi
对于输入框，您可以检查此链接


[ ^ ]

在每个复选框更改事件上，您都可以编写此代码以获取输入

hi
for input box u can check this link


[^]

and on every checkbox changes event you can write this code for getting input

您需要的是查看单选按钮的选择内容，然后发布一个对话框，以防单选按钮被选中"可怜".

您可以将onchange属性绑定到单选按钮以进行检查.
下一篇文章提供了有关如何执行此操作的示例:使用JavaScript验证单选按钮 [

What you need is to see what was selected for the radio button and then post a dialog box in case radio button selection was ''poor''.

You can bind an onchange property to radio button to check the same.
Following article has an example on how to do it: Using JavaScript to Validate Radio Buttons[^]

There are more samples on web if needed. I already shared the logic, try out.

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