Javascript jquery下拉列表和发布到sql数据库问题

问题是我们无法让它从下拉框中选择信息并将其发布到我们的数据库中。我们尝试了许多看不见的方法，以获取所选下拉列表的值，使其显示在任何位置。很抱歉，第一次发布堆栈溢出

HTML和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; ?>"});
});

尝试：

而不是这一行：

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

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

您还需要在这一行中输入：

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

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

尝试：

而不是这一行：

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

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

您还需要在这一行中输入：

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

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