Fel — webForum

Alpha IIMedlem sedan maj 20002 993 inlägg

4 juni 2001 20:23#1

Varför fungerar inte det här?

Warning: Supplied argument is not a valid MySQL result resource in C:\windows\Skrivbord\LastUpdates\login.php on line 8

<?php
 session_start();
 session_register('username');
 session_register('password');

if(IsSet($stage)) {
 $result = mysql_query("SELECT * FROM administrators WHERE username = '$login_name' and password = '$login_pass'");
 $found = mysql_num_rows($result);
 if($found) {
   $admin_info = mysql_fetch_row($result);
   $username = $admin_info[1];
   $password = $admin_info[2];
   header("Location: index.php");
 }
}
?>
<html>
<head>
 <title>Latest Updates</title>
<style type="text/css">
  body { background: #ffffff; color: #000000; }
  .header { font-family: Verdana, Arial; font-size: 11px; color: #000000; font-weight: bold;}
  .text1 { font-family: Verdana, Arial; font-size: 11px; color: #000000; }
  .text2 { font-family: Verdana, Arial; font-size: 9px; color: #000000; }

  a:link { color: #336699; }
  a:visited { color: #56A9D0; }
  a:active { color: #56A9D0; }
</style>
</head>

<body>
<table border="0" width="400" cellpadding="0" cellspacing="5">
  <tr>
    <td height="60"><img src="img/loggo.gif" border="0"></td>
  </tr>
  <tr>
    <td>
      <table border="0" width="400" cellpadding="0" cellspacing="0">
        <tr>
          <td colspan="3"><img src="img/roundedtop.gif" border="0"></td>
        </tr>
        <tr>
          <td width="1" bgcolor="#00000"><img src="img/empty.gif" border="0"></td>
          <td width="398">
            <table border="0" width="398" cellpadding="5" cellspacing="0">
              <tr>
                <td>
                  <p class="header">Login</p>

<?php
 if(IsSet($stage))
   echo("<p class=\"text1\"><font color=\"red\"><b>Fel lösenord/användarnamn!</b></font></p>");
?>

                  <p class="text1">
                    <form method="post" action="login.php">
                    Användarnamn:<br>
                    <input type="text" name="login_name"><br>
                    Lösenord:<br>
                    <input type="password" name="login_pass"><br>
                    <input type="hidden" name="stage" value="1">
                    <input type="submit" value="Login">
                    </form>
                  </p>

                  <p class="text2" align="right">&copy; <a href="mailto:daniel.henell@telia.com">Daniel Henell 2001</a></p>
                </td>
              </tr>
            </table>
          </td>
          <td width="1" bgcolor="#00000"><img src="img/empty.gif" border="0"></td>
        </tr>
        <tr>
          <td colspan="3"><img src="img/roundedbottom.gif" border="0"></td>
        </tr>
      </table>
    </td>
  </tr>
</table>
</body>
</html>

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I have not failed. I´ve just found 10.000 ways that won´t work. // Thomas Edison

SPiNMedlem sedan mars 20007 896 inlägg

4 juni 2001 21:23#2

Ser rätt ut... Prova följande istället för rad 8 och 9:

$found = mysql_affected_rows();
if($found > 0) {...

Vet inte om det gör nån skillnad, men prova :)

------------------
SPiN, bjorne.w@telia.com

-- They pretend to pay me, I pretend to work --

LarsGMedlem sedan dec. 200012 464 inlägg

4 juni 2001 21:53#3

Lägg in lite felhantering

$result = mysql_query("SELECT * FROM administrators WHERE username = '$login_name' and password = '$login_pass'") 
or die(mysql_error());

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essentitia preter non sans multiplicandum

Alpha IIMedlem sedan maj 20002 993 inlägg

5 juni 2001 20:46#4

Hade glömt att connecta db:n... klantigt

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I have not failed. I´ve just found 10.000 ways that won´t work. // Thomas Edison