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개발자팁

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애왜안될까요;;; 정보

MySQL 애왜안될까요;;;

본문

<?
include "dbcon.php";
if($num == "")
{popup_back("잘못된 경로 입니다.");}

$sql = "select * from message wgere num = $num";
$res = mysql_query($sql);

show();

mysql_close($con);
//--------------show()함수--------------//
function show()
{
global $res;

$num = mysql_result($res, 0, 'num');
$name = mysql_result($res, 0, 'name');
$email = mysql_result($res, 0, 'email');
$message = mysql_result($res, 0, 'message');
$writedate = mysql_result($res, 0, 'writedate');
$sourceip = mysql_result($res, 0, 'sourceip');

echo("
<table border=1>
  <tr>
   <td>번호</td>
   <td>이름</td>
   <td>이메일</td>
   <td>작성일</td>
   <td>IP</td>
  </tr>
  <tr>
   <td colspan=5>메시지</td>
  </tr>
  </table>
");
}
//--------------show()함수--------------//
?>

이렇게 했거든여

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in C:\AutoSet6\public_html\view.php on line 18

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in C:\AutoSet6\public_html\view.php on line 19

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in C:\AutoSet6\public_html\view.php on line 20

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in C:\AutoSet6\public_html\view.php on line 21

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in C:\AutoSet6\public_html\view.php on line 22

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in C:\AutoSet6\public_html\view.php on line 23
이런 에러가나네요

mysql연결은 다되있고 확인까지 했는데 저러네용;;