jquery下拉併發布到sql數據庫問題

Question

問題是我們不能從下拉框中選擇信息並將其發佈到我們的數據庫中。我們嘗試了許多事情，無法看到選定的下拉列表的值顯示在任何地方。對不起，第一次發佈堆棧溢出。

HTML和PHP的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> 

的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; ?>"}); 
}); 

Answer 1

嘗試：

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

，而不是這一行：

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

您還需要該行上的（）：

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