jQuery clone（）无法按预期工作

Question

Html code:

<form>
    <table>
        <tr>
            <td>Value</td>
            <td class="to">ABC</td>
        </tr>
    </table>
</form>
<div class="from" style="display:none;">
    <label>
        <input type="text" value="" placeholder="Number"/>
    </label>
</div>

and this javascript code:

$(document).ready(function(){
          var values = [12,11,15];
          var cloned = $('div.from').find('label').clone();
          $('td.to').empty();
          for(var i=0;i<values.length;i++){
              cloned.find('input').val(values[i]);
              //console.log(cloned);
              console.log(cloned.find('input').val());
              $('td.to').append(cloned);
          }
});

I am here trying to clone the nested ' label ' from the hidden div ' div.from ' and set one of the value from the ' values ' array and append to empty ' td.to ' using the same cloned object. As far as I know, the last value of the array seems to be appended after cloning. The result should have been three cloned nested labels with value populated in the input box from the ' values ' array.

Here is jsbin link : http://jsbin.com/yuyaqu

Answer 1

You need to clone() once per iteration, so put that line within the for :

var values = [12,11,15];
$('td.to').empty();
for (var i = 0; i < values.length; i++){
    var cloned = $('div.from').find('label').clone();
    cloned.find('input').val(values[i]);
    $('td.to').append(cloned);
}