如果無線輸入檢查顯示下一個問題

Question

#Jquery

 $("input").on('change ', function() { 

     var ac_field = $("fieldset .active").children("input[type='radio']:checked"); 

     if(ac_field){ 
      var next = ac_field.next("fieldset"); 

      ac_field.addClass("unactive"); 
      ac_field.removeClass('active'); 

      next.addClass("active"); 
      next.removeClass('unactive'); 
     } 

    }) 

CSS

.active{ 
     display: block; 
    } 
    .unactive{ 
     display: none; 
    } 

HTML

 <!--Q_C1--> 
    <fieldset class="Q_C_1 active"> 
     <legend class="q_holder"></legend> 

     <div class="c_cont"> 

      <div class="C_A C_A_mI"> 
       <input type="radio" id="C_1_1" class="rad" name="Q_C_1" value="1"> 

       <label for="C_1_1" class="too_unhappy face_fix_all"></label> 
      </div> 

      <div class="C_A"> 
       <input type="radio" id="C_1_2" class="rad" name="Q_C_1" value="2"> 

       <label for="C_1_2" class="unhappy face_fix_all"></label> 
      </div> 

      <div class="C_A"> 
       <input type="radio" id="C_1_3" class="rad" name="Q_C_1" value="3"> 

       <label for="C_1_3" class="nut face_fix_all"></label> 
      </div> 

      <div class="C_A"> 
       <input type="radio" id="C_1_4" class="rad" name="Q_C_1" value="4"> 

       <label for="C_1_4" class="happy face_fix_all"></label> 
      </div> 

      <div class="C_A C_A_mII"> 
       <input type="radio" id="C_1_5" class="rad" name="Q_C_1" value="5"> 

       <label for="C_1_5" class="too_happy face_fix_all"></label> 
      </div> 

      <span class="stretch"></span> 

     </div> 

    </fieldset> 




    <!--Q_C2--> 
    <fieldset class="Q_C_2 unactive"> 
     <legend class="q_holder"></legend> 

     <div class="c_cont"> 

      <div class="C_A C_A_mI"> 
       <input type="radio" id="C_2_1" class="rad" name="Q_C_2" value="1"> 

       <label for="C_1" class="too_unhappy face_fix_all"></label> 
      </div> 

      <div class="C_A"> 
       <input type="radio" id="C_2_2" class="rad" name="Q_C_2" value="2"> 

       <label for="C_2_2" class="unhappy face_fix_all"></label> 
      </div> 

      <div class="C_A"> 
       <input type="radio" id="C_2_3" class="rad" name="Q_C_2" value="3"> 

       <label for="C_2_3" class="nut face_fix_all"></label> 
      </div> 

      <div class="C_A"> 
       <input type="radio" id="C_2_4" class="rad" name="Q_C_2" value="4"> 

       <label for="C_2_4" class="happy face_fix_all"></label> 
      </div> 

      <div class="C_A C_A_mII"> 
       <input type="radio" id="C_2_5" class="rad" name="Q_C_2" value="5"> 

       <label for="C_2_5" class="too_happy face_fix_all"></label> 
      </div> 

      <span class="stretch"></span> 

     </div> 

    </fieldset> 

我想它來檢查，如果輸入[單選]是檢查

如果有任何<fieldset>（1）元素具有子輸入[無線電]檢查

然後獲得下<fieldset>（2）設置其類活性（顯示：塊）

和<fieldset>（1）爲不活動（顯示：無）

（是無線電輸入檢查顯示下一個問題） 我有10 <fieldset>

Answer 1

使用這個劇本，也有在路上幾個錯誤，您正在使用的選擇

$("input").on('change ', function() { 
    var ac_field = $("input[type='radio']:checked").closest("fieldset.active");//changes here 

    if(ac_field){ 
     var next = ac_field.next("fieldset"); //changes here 

     ac_field.addClass("unactive"); 
     ac_field.removeClass('active'); 

     next.addClass("active"); //changes here 
     next.removeClass('unactive'); //changes here 
    } 
});