Question

我有一个jQuery星评级。我的要求是点击按钮获取星号值。

这是我的代码

<div id="Ratingdiv">
    <input type="radio" class="star" name="dynamic-control" value="1" />
    <input type="radio" class="star" name="dynamic-control" value="2" />
    <input type="radio" class="star" name="dynamic-control" value="3" />
    <input type="radio" class="star" name="dynamic-control" value="4" />
    <input type="radio" class="star" name="dynamic-control" value="5" />
</div>

<div style="padding: 20px">
    <form name="html-test" id="html-test">
    <input type="button" value="Insert control"      onclick="insert_dynamic_control(this.form);" />
    <div class="controls" style="padding: 5px; clear: both;">
</div>
<div style="padding: 5px; clear: both;">
     <input id="Button1" type="button" onclick="Getval();" value="button" />
    <span></span>
</div>
</form>
</div>

点击时使用的jQuery是

     <script type="text/javascript">
       {
          function Getval() {
              (function($) {
              $().ready(function() {
                var main = $('#Ratingdiv');
                main.each(function() {
                    var odds = $(main).find('.star');

                    alert(odds);

                });
            });
        })(jQuery);
    }

我在警告框中未定义。请帮助我

Answer 1

检查演示

http://jsfiddle.net/sandeepscet/NEEEB/2/

$('#Button1').click(function(){

   alert( $('input:radio[name=dynamic-control]:checked').val());


});

Answer 2

http://jsfiddle.net/xJqtB/

$(function() {
    $('#Button1').on('click', function(e){
        var main = $('#Ratingdiv');
        main.each(function() {
            var odds = $(main).find('.star:checked').val();

            alert(odds);
        });
    });
});

jQuery评级不起作用

2 个答案: