从select选项标签定义要创建的数据库

html表单动作属性=" uploadone.php"。表单有一个选项元素value attribute = 'Red'。有许多数据库将以这种方式创建。我可以像下面这样给URL添加参数，但是因为选择很长，所以太麻烦了。这种情况下的最佳解决方案是什么?请在这方面提供帮助。由于前期

 <form action="uploadone.php" method="post"                
 enctype="multipart/form-data">
 <select name="unoone" class="unoone ui-btn ui-mini" data-role="none"     data-native-menu="false" required>
  <option value="">My Task</option>
  <option option value="Red">Red</option>
  <option option value="Blue">Blue</option>
  <option option value="Black">Black</option>
  <option option value="White">White</option>    
 <input type="submit" value="Submit" name="submit" />
  </form>

uploadone.php

  include 'Includes/conDB.php';
 // 'Red' the option element value attribute
 // Select a database to use 
 // T1 and T2 have a LEFT JOIN
 if ($_POST['unoone'] == 'Red') {
include "Includes/T1.php"; 
include "Includes/T2.php";
}