我试着做一个动态下拉，当在医生下拉列表中选择任何多边形选项时，总是出现同一个医生。
这是动态下拉的代码

<div class="form-group">
            <label>Poly : </label>
            <select name="poly" id="poly" class="form-control" style="width: 50%">
                <option disabled selected>--Choose Poly--</option>
                <?php
                $sqql = mysqli_query($kon,"SELECT * FROM poly") or die(mysqli_error($kon));
                while ($rorow = mysqli_fetch_array($sqql)) {
                    ?>
                    <option value="<?php $rorow['id_poly'];?>"><?php echo $rorow['poly']; ?></option>
                    <?php
                }
                ?>
            </select>
        </div>
        <div class="form-group">
            <label>Doctor : </label>
            <select name="doctor" id="doctor" class="form-control" style="width: 50%">
                <option disabled selected>--Choose doctor--</option>
            </select>
        </div>

这是ajax的代码

<script   src="https://code.jquery.com/jquery-3.1.1.js"   integrity="sha256-16cdPddA6VdVInumRGo6IbivbERE8p7CQR3HzTBuELA="   crossorigin="anonymous">  </script>
<script type="text/javascript">
    $(document).ready(function(){
        $("#poly").change(function(){
            var id_poly = $(this).val();
            $.ajax({
                url:"getdoctor.php",
                method:"POST",
                data: {id_poly:id_poly},
                success:function(data)
                {
                    $("#doctor").html(data);
                }
            });
        });
    });
</script>

这是getdoctor.php的代码

$ouput ='';
$dok = mysqli_query($kon,"SELECT * FROM doctor WHERE id_poly = '".isset($_POST["id_poly"])."'") or die(mysqli_error($kon));
$output .= '<option value="" disabled selected>--Choose doctor--</option>';
if (mysqli_num_rows($dok)>0){
    while ($row = mysqli_fetch_array($dok)) {
        $output .= '<option value="'.$row["id_doctor"].'">'.$row["doctor"].'</option>';
    }
}
echo $output;