javascript表单验证，其中不包括隐藏的下拉菜单

Question

The code for my project is http://jsfiddle.net/CVS2F/2/ . I also have a submit button which submits the data to mysql, but when i click attending: 'no' there is no value submitted by user as they are not attending dropdowns are hidden but with the code an error pops up anway. is there a way around this to ignore hidden dropdowns?

Answer 1

In the php code add these lines:

if($_REQUEST['yesNo'] == 'yes)
    {
    // form process with list
    }
    else
    {
    // form process without list
    }