如何讓從選定字段中選擇的值不再消失？

Question

我有以下代碼：

<form method="post" enctype="multipart/form-data"> 
<table id="box-table-a" summary="PAGE HEADER" style='width:10%'> 
<tr> 
<th> 
<?php 
include('./db.php'); 
$PM = mysqli_query($con, "SELECT DISTINCT Name FROM report ORDER BY Name ASC"); 
echo "<b>Sort by Name:</b>&nbsp;\n"; 
echo " <select name='Name' onChange='submit(this.form)'>\n"; 
while($row = mysqli_fetch_row($PM)) { 
$sel = ($table === $row[0]) ? "id='sel' selected" : ""; 
printf(" <option %s value='%s'>%s</option>\n", $sel, $row[0], $row[0]); 
    } 
    echo " </select>\n"; 
    ?> 
</th> 
</tr> 
</form> 

一旦做出選擇，選擇框默認回任何值是在列表的頂部（從SELECT SQL查詢...）。如何阻止這種情況的發生並保持在選擇框中選定的值，直到做出其他選擇？