mysql - Dynamically alter drop down by the results of selecting another dynamic drop down php -

i trying populate second drop down results of first drop down box. want them dynamic mysql tables. have far, second drop down wont populate. based on results of selecting project_id, want see tasks under selected project. appreciated!!!! yes aware of mysql_query , vulnerabilities. intend on going after works correctly , fixing security issues.

there 3 tables involved projects, employee, , tasks.

projects:  project_id, project_name    tasks:  task_id, project_id employee:  staff_id, lname, fname 

code follows:

<?php require_once('mainheader.php'); require_once('db_credentials.php'); ?>  <!-- latest compiled , 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 , minified javascript --> <script src="//netdna.bootstrapcdn.com/bootstrap/3.1.1/js/bootstrap.min.js"></script>     <div class="container">  <h2>please fill out fields 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());  $projsql = mysql_query("select * projects activeproject ='open'") or die(mysql_error());  while($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 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());  $tasksql = mysql_query("select * tasks project_id=" .$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 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());      $empsql = mysql_query("select *  employee")     or die(mysql_error());      while($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 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'); ?>