jquery下拉列表并发布到sql数据库问题

The issue is that we cant get it to select the information from the dropdown box and post it into our database. We tried numerous things cant see to get the value of the selected dropdown to display anywhere. Sorry first time posting on stack overflow.

HTML and PHP Dropbox

<div class="tl">
  <center><br><? $sqltl = "SELECT * FROM till WHERE account ='$dbname' ORDER BY tillname ASC";
    $resulttl = mysql_query($sqltl);

    echo "<select id='tills' name='tills'>";
    while ($rowtl = mysql_fetch_array($resulttl)) {
      echo "<option value='" . $rowtl['tillname'] . "'>" . $rowtl['tillname'] . "</option>";
    } echo "</select>"; ?><br>

    <div class="pagination btn-group">
      <div class="btn btn-medium highlight-color-0" id="floattl" style="width: 150px; height: 150px; margin: 0.5px; white-space: normal">
        <div class="btn-image dollar-bill"></div>
        <span class="btn-text">Float Till</span>
      </div>
      <div class="btn btn-medium highlight-color-0" id="closetl" style="width: 150px; height: 150px; margin: 0.5px; white-space: normal">
        <div class="btn-image dollar-bill"></div>
        <span class="btn-text">Close Till</span>    
      </div>
    </div>
  </center>

  <div>Current Float Amount</div>
    <div class="btn btn-small highlight-color-0" id="submit_float">Confirm</div>
    <input type="text" id="tlfloat_num" style="width: 275px;">
  </div>

jQuery:

$('#floattl').click(function() {
  tillselect = $( "#tills option:selected" ).val;
  $('#tlfloat_num').val(tillselect);
});

$('#submit_float').click(function() {
  var tillfloat = $('#tlfloat_num').val;
  var $openingamount = tillfloat;
  //var $select_tilname = $('#tills').get(0).selectedIndex = 0;

  //$('#tlfloat_num').val($openingamount);
  //$('#tlfloat_num').val($tillselect);

  $.post("till.php", {tlname: tillselect, openingamount: $openingamount, account: "<? echo $dbname; ?>"});
});

try:

var tillselect = $('#tills option:seclected').val();
                                                 ^^^

instead of this line:

tillselect = $( "#tills option:selected" ).val;

You also need the () on this line:

var tillfloat = $('#tlfloat_num').val;
//var tillfloat = $('#tlfloat_num').val();