使用jquery關閉模式彈出框

Question

下面的代碼是我的彈出視圖...我想關閉彈出窗口上點擊關閉按鈕（X）這是在彈出式頂部...在視圖的底部我有jquery關閉popup..but那不是working..i有像父母彈出和兒童的另一個popup..just彈出popup.and在這種情況下，我不得不關閉彈出的孩子......請幫我

<div class="modal-dialog" style="width:1056px" id="pop"> 
    <div class="modal-content"> 
     <div class="modal-header">    
      <input type="button" onclick="close()" id="button" value="Close [x]" />   

     </div> 
     <div class="modal-body"> 
      <div class="scroller"> 
       <table class="display table table-striped table-bordered table-hover" id='tblTicket' cellspacing="0" width="100%"> 
        <thead> 
         <tr> 
          <th>Data</th>        
         </tr> 
        </thead> 
        <tbody> 
         <tr> 
          <td>Data</td>        
         </tr> 

        </tbody> 
       </table> 


       <table class="display table table-striped table-bordered table-hover" id='tblTicket' cellspacing="0" width="100%"> 
        <tr> 
         <th>DATA</th>       
        </tr> 

        @foreach (var item in Model) 
        { 
         <tr> 
          <td> 
          data 
          </td>       
         </tr> 
        } 

       </table> 
      </div> 
     </div> 
    </div> 
</div> 

<script type="text/javascript" language="javascript"> 
$("#button").click(function() {   
    $("#pop").close 
     }); 
</script> 

Answer 1

嘗試這

<script type="text/javascript" language="javascript"> 
$("#button").click(function() {   
    $('#pop').modal('hide'); 
     }); 
</script> 

Answer 2

請檢查下面的代碼：

<script type="text/javascript" language="javascript"> 

    $("#button").click(function() { 
     $('#pop').modal('toggle'); 
     $('#pop').modal('hide'); 
     $('.modal-backdrop').removeClass('modal-backdrop'); 
     $('.fade').removeClass('fade'); 
     $('.in').removeClass('in'); 

     $('html, body').css({ 
      'overflow': 'auto', 
      'height': 'auto' 
     }); 

    }); 

</script> 

乾杯！

Answer 3

試試這個代碼： -

$("#button").click(function() { 

    $('#pop').modal('toggle'); 

});