php소스를 asp소스를 바꿀수 있을까염?

아래 소스를 asp소스로 바꾸고 싶어요.. 도와주세요!! 어떻게 바꿔야 할지 막막하네요... 도와주세요!! 네?.....ㅡㅜ <HTML> <BODY background=bg.gif> <center> <img src=2.gif><img src=title.gif><img src=2.gif> <? //bgproperties="fixed" $board="naboard"; $db = mysql_connect("localhost","galgu","--"); mysql_select_db("galDB",$db); $day2=(int)date("d"); $value = mysql_query("delete from $board where dayofmonth(wdate)! = $day2",$db); $result = mysql_query("select * from $board order by wdate desc limit 12",$db); $a=0; echo "<TABLE border=0 WIDTH=600>"; while($row=mysql_fetch_array($result)) { if($a == 0 || $a == 4 || $a == 8 || $a == 12) { print "<tr>"; } print "<td width=20%>"; //$content=htmlspecialchars($row[content]); //$content=str_replace("\n"," ","$content"); //$content=str_replace(" ","&nbsp;",$content); $content = $row[content]; $content =str_replace("\n"," ","$content"); $name = htmlspecialchars($row[name]); echo (" <table border=0> <tr> <td colspan=2>$name</TD></TR> <TR> <TD></TD> <TD>$content</td></tr> <tr> <td colspan=2>$row[u_ip]</td></tr> </table> "); print "</td>"; if($a==3 || $a==7 || $a ==11 || $a ==15) {print "</tr>"; } $a++; } echo "</TABLE>"; mysql_close($db); ?> <A HREF = "write.php" onMouseOver = "document.images['imBut'].src='go1.gif'" onMouseOut = "document.images['imBut'].src='go.gif'"> <IMG NAME="imBut" SRC="go.gif" BORDER=0 alt="낙서 안하고 뭐해 ??"></A> </CENTER> </BODY> </HTML>