2014-01-19 106 views
0

请参阅图像我正在尝试执行的操作我在每个td中都有HTML表格我有下拉菜单和复选框,当我在下拉列表中选择值时我想要使复选框检查特定的td任何人都可以指导我如何做到这一点。如何在选择下拉菜单时选中该复选框

enter image description here

代码:

<td style="width:141px" > <select name="mySelect" id="mySelect" class="edit1 route '.$rows["net_id"].'" >   
      <option value="-1">Choose..</option>'; 


      $query = "SELECT route FROM routestable WHERE `clientid` = '$client_id' "; 

      $result = mysql_query($query); 

      while ($rows1 = mysql_fetch_assoc($result)) 
      { 


        if(strlen($rows1['route'])>0 && $rows1['route']==$rows['route']){ 
    print' <option value="'.$rows1['route'].'" selected>'.$rows1['route'].' </option>';} 

       else{ 

       echo '<option value="' . $rows1['route'] . '"> ' . $rows1['route'] .  '</option>'; 

       } 

      } 


     echo '</select> 
          </td> 


        <td style="width:200px" id="CPH_GridView1_Status1'.$rows['net_id'].'" class="edit2 status1 '.$rows["net_id"].' "><input type="checkbox" style="margin:0 0 0 93px;" name=""/></td> 

AJAX

<script> 

    $(document).ready(function(){ 

     $('.edit1').on('change', function(){ 


            arr = $(this).attr('class').split(" "); 
            var clientid=document.getElementById("client").value; 
            account_id=document.getElementById("account_id").value; 


             $.ajax({ type: "POST", 
                url:"clientnetworkpricelist/routestatusupdate.php", 
                data: "value="+$(this).val()+"&rowid="+arr[2]+"&field="+arr[1]+"&clientid="+clientid+"&account_id="+account_id, 


                success: function(res){ 
                 data = jQuery.parseJSON(res); //added line 
                 alert('Saved Successfully!'); 
                 $('#CPH_GridView1_Status'+arr[2]).empty(); 
                 $('#CPH_GridView1_Status'+arr[2]).append(data.status); 
                 $('.ajax').html($(this).val()); 
                 $('.ajax').removeClass('ajax'); 
                }}); 


            } 


         ); 


    }); 


</script> 
+1

在ajax请求之前的变更句柄中,尝试'$(this).closest('td')。next()。find('input')。prop('checked',true) –

回答

2

要选中复选框,在未来的TD试

$(this).parent().next().find('input:checkbox').attr("checked",true); 

Jsfiddle