第二个下拉框取决于我的第一个下拉框.如何去做 - Second drop down depends on my first drop down. How?

Second drop down depends on my first drop down. How?

我的程序中有两个下拉菜单，我希望它是这样的:

我有一个员工列表，并显示/填充列表并放置它在我的第一个下拉菜单中。
第二个是，当我选择"Francis"时，第二个下拉框将当我选择另一个雇员时，只有Francis才有一个值/事件。

我认为第一个下拉是正确的，但我想得到它的值，并把它放在我的SQL语句中的WHERE子句在我的第二个下拉。如何?

<tr>
    <td><label for="cname">Client Name:</label></td>
        <td><select name="cname" id="cname">
            <option>Choose</option>
            <?php
                include("alqdb.php");
                $result=mysqli_query($con, "SELECT ClientName FROM events");
                    while($row=mysqli_fetch_assoc($result)){
                        echo "<option>".$row["ClientName"]."</option>";
                    }
            ?>
            </select>
        </td>
    </tr>
    <tr>
        <td><label for="survey">Survey:</label></td>
        <td><select name="survey" id="survey">
            <option>Choose</option>
            <?php
                include("alqdb.php");
                $result=mysqli_query($con, "SELECT EventTitle FROM events WHERE ClientName = 'Francis'");
                    while($row=mysqli_fetch_assoc($result)){
                        echo "<option>".$row["EventTitle"]."</option>";
                    }
            ?>
            </select>
        </td>
    </tr>

为第二个select执行ajax请求:

为ajax请求创建PHP文件html:

<tr>
    <td><label for="cname">Client Name:</label></td>
        <td><select name="cname" id="cname">
            <option>Choose</option>
            <?php
                include("alqdb.php");
                $result=mysqli_query($con, "SELECT ClientName FROM events");
                    while($row=mysqli_fetch_assoc($result)){
                        echo "<option>".$row["ClientName"]."</option>";
                    }
            ?>
            </select>
        </td>
    </tr>
    <tr>
        <td><label for="survey">Survey:</label></td>
        <td><select name="survey" id="survey">
            <option>Choose</option>
        </select>
        </td>
    </tr>

ajax.php

 <?php
     include("alqdb.php");
     $result=mysqli_query($con, "SELECT EventTitle FROM events WHERE ClientName = '"$_GET['cname']"'");
     while($row=mysqli_fetch_assoc($result)){
         echo "<option>".$row["EventTitle"]."</option>";
     }
 ?>

js:

<script>
$(function(){
 $('#cname').on('change',function(){
    $.ajax({
     url:ajax.php,
     data:{cname:$('#cname').val()};
     type:'get',
     contentType:'html',
     success:function(data){
     $('#survey').append(data);
    }
    });
});
});
</script>