Php 通过选择另一个动态下拉菜单的结果动态改变下拉菜单

我试图用第一个下拉框的结果填充第二个下拉框。我希望它们在mysql表中是动态的。这是我到目前为止所做的，第二个下拉列表不会填充。根据选择项目id的结果，我只想查看所选项目下的任务。我们将感谢所有的帮助！！！！是的，我知道mysql\u query及其漏洞。我打算在一切正常工作并解决所有安全问题后返回

涉及的三个表是

项目

、

员工

和

任务

projects:  project_id, project_name
   tasks:  task_id, project_id
employee:  staff_id, lName, fName

代码如下：

<?php
require_once('mainheader.php');
require_once('db_credentials.php');
?>

<!-- Latest compiled and minified CSS -->
<link rel="stylesheet" href="//netdna.bootstrapcdn.com/bootstrap/3.1.1/css/bootstrap.min.css">

<!-- Optional theme -->
<link rel="stylesheet" href="//netdna.bootstrapcdn.com/bootstrap/3.1.1/css/bootstrap-theme.min.css">

<!-- Latest compiled and minified JavaScript -->
<script src="//netdna.bootstrapcdn.com/bootstrap/3.1.1/js/bootstrap.min.js"></script>
    <div class="container">
 <h2>Please fill out all fields to register new Customers</h2>
    <form action="task_assignment_handler.php" method="post"> 
    <form class="form-vertical" role="form">

<?php

mysql_connect( $db_hostname, $db_username, $db_userpass ) or die(mysql_error());
mysql_select_db( $db_dbname ) or die(mysql_error());

$projsql = mysql_query("SELECT * FROM projects WHERE activeproject ='open'")
or die(mysql_error());

while($ProjectResults = mysql_fetch_array( $projsql ))
{
    $proj_id = $ProjectResults['project_id'];
    $proj_name = $ProjectResults['project_name'];
    $project_block .= '<OPTION value="'.$proj_id.'">'.$proj_name.'</OPTION>'; 
} 

?>
    <div>
    <label for="customer">Select a Project</label>
    <select name="task_select"><?php echo $project_block; ?></select>
    </div>




    <?php

mysql_connect( $db_hostname, $db_username, $db_userpass ) or die(mysql_error());
mysql_select_db( $db_dbname ) or die(mysql_error());

$tasksql = mysql_query("SELECT * FROM tasks WHERE project_id=" .$proj_id)
or die(mysql_error());

while($TaskResults = mysql_fetch_array( $tasksql ))
{

    $task_id = $TaskResults['task_id'];
    $task_name = $TaskResults['taskname'];
    $task_block .= '<OPTION value="'.$task_id.'">'.$task_name.'</OPTION>'; 
} 

?>
    <div>
    <label for="customer">Select Task to assign</label>
    <select name="task_select"><?php echo $task_block; ?></select>
    </div>

<?php

    mysql_connect( $db_hostname, $db_username, $db_userpass ) or die(mysql_error());
    mysql_select_db( $db_dbname ) or die(mysql_error());

    $empsql = mysql_query("SELECT *  FROM employee")
    or die(mysql_error());

    while($empresults = mysql_fetch_array( $empsql ))

    {
    $employee_id = $empresults['staff_id'];
    $employee_fname = $empresults['fName'];
    $employee_lname = $empresults['lName'];
    $employee_block .= '<OPTION value="'.$employee_id.'">'.$employee_fname.'&nbsp'.$employee_lname.'</OPTION>'; 
    } 

?>

    <div>
    <label for="employee">Assign Employee to Task</label>
    <select name="employee_select"><?php echo $employee_block; ?></select>
    </div>










    <button type="submit" class="btn btn-default">Assign Task</button>
    </div>


    <?php 
require_once('mainfooter.php');
?>

请填写所有字段以注册新客户
使用ajax
。。。