javascript - jquery dropdown and posting to sql database issue -

the issue cant select information dropdown box , post our database. tried numerous things cant see value of selected dropdown display anywhere. sorry first time posting on stack overflow.

html , php dropbox

<div class="tl">   <center><br><? $sqltl = "select * till account ='$dbname' order 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 line:

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

you need () on line:

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