simple logon script using DB

[Anon]

hi,

i know that you don't really like correcting code but i'm hoping you can do that for me as i can't get this simple script to work. It uses the same code to access the database as my others and they work fine.

Here's my code which at the moment gives this error:

Supplied argument is not a valid MySQL result resource in ........

<?
/* DB Variables */
$hostname = "";
$username = "";
$password = "";
$dbName = "";
$userstable = "users";

include "header.php3";

/* login */
if (!$login)
 {
?>
<p align="center" class="newsheadline"><br>
Members Only<br>
<br>
Login<br>
</p>

<form action="membersonly2.php3" method="POST">
  <div align="center"><center><table border="0" width="80%" cellspacing="0">
    <tr>
      <td width="50%" class="newstext"><div align="left"><p><b>Username:</b></td>
      <td width="50%"><input TYPE="text" NAME="usrname" SIZE="20" MAXLENGTH="30"></td>
    </tr>
    <tr>
      <td width="50%" class="newstext"><b>Password:</b></td>
      <td width="50%"><input TYPE="password" NAME="passwd" SIZE="20" MAXLENGTH="30"></td>
    </tr>
    <tr>
      <td width="50%"></td>
      <td width="50%"><input TYPE="submit" name="login" VALUE="Login"></td>
    </tr>
  </table>
  </center></div>
</form>
<?
include "footer.php3";
 }
else 
 {
/* make connection to database */
MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database");

@mysql_select_db( "$dbName") or die( "Unable to select database"); 

/* Select all users */
$query = "SELECT * FROM $userstable WHERE $userstable.user = '$usrname'";

$result = MYSQL_QUERY($query);

$number = MYSQL_NUMROWS($result);

$i = 0;

if ($number == "0") 
	{
	/* Print error */
	PRINT "<CENTER>";
	PRINT "<br><font class=newstext><big>ERROR: Invalid username.</big><br>Please use your browser back button and try again.";
	PRINT "</CENTER>";
	include "footer.php3";
	}
else
	{
		$user = mysql_result($result,$i,"user");
		$pass = mysql_result($result,$i,"pass");
		$level = mysql_result($result,$i,"level");
	if ($passwd == "$pass")
	 {
?>
<p align="center" class="newsheadline"><br>
Members Only: News
</p>
<p align="center" class="newstext">
Welcome <? echo("$usrname"); ?> to our members only area.<br>
</p>
<?
include "mnews.txt";
include "footer.php3";
 }
else
 {
/* Print error */
PRINT "<CENTER>";
PRINT "<br><font class=newstext><big>ERROR: Invalid password.</big><br>Please use your browser back button and try again.";
PRINT "</CENTER>";
include "footer.php3";
 }
}
}
?>

I'm fairly sure it's a simple error so if you could help me out i'd be very grateful.

Dan

[Anon]

Hey Dan!

If I were you, first of all i would try the mysql_error([db link identifier]) function, which tells you exactly what the mysql error is caused by.

Just a quick look at your code, tells me that this part of the code is not optimal:

else
{
$user = mysql_result($result,$i,"user");
$pass = mysql_result($result,$i,"pass");
$level = mysql_result($result,$i,"level");

When accessing the variable in which data returned from the mysql query, I usually use "$row = mysql_fetch_array($query_result)". Then you can access each variable int the array returned in $row, by typing $row[usrname] for example.

I hope that was of some help... - Mikkel

[Anon]

thanks for the reply but i don't think there is anything wrong with the query but there's something else very simple that's not right but i just can't find it!

[galmar]

I see an @ in front of mysql_select_db