重新加载页面的php脚本

我对不能立即显示更新数据的页面有困难 这是我现在的一个:

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("hospital", $con);

$result = mysql_query("SELECT * FROM t2");

 echo"<br>";
 echo"<big>All In Patients</big>";
echo "<table border='1'>
<tr>
<th>Pnum</th>
<th>HospNum</th>
<th>RoomNum</th>
<th>LastName</th>
<th>FirstName</th>
<th>MidName</th>
<th>Address</th>
<th>TelNum</th>
<th>Stat</th>
<th>Nurse</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
   echo "<tr>";
   echo "<td>" . $row['PNUM'] . "</td>";
  echo "<td>" . $row['HOSPNUM'] . "</td>";
  echo "<td>" . $row['ROOMNUM'] . "</td>";
    echo "<td>" . $row['LASTNAME'] . "</td>";
      echo "<td>" . $row['FIRSTNAME'] . "</td>";
        echo "<td>" . $row['MIDNAME'] . "</td>";
          echo "<td>" . $row['ADDRESS'] . "</td>";
           echo "<td>" . $row['TELNUM'] . "</td>";
            echo "<td>" . $row['STAT'] . "</td>";
            echo "<td>" . $row['NURSE'] . "</td>";

  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>

<form>
<input type="button"  class='type_button'  value="Print" onClick="window.print();" /> 
</form>
</body>
</html>

你知道有什么办法可以更新上面的代码吗?因为当我更新数据然后加载上面的代码时。它将只显示以前尚未更新的数据。您必须右键单击并刷新页面才能看到效果。