[SOLVED] php help_connecting sql database

[ma5ect]

Hi,

 

I 'am trying to connect my sql database with my webpage for users log in. i have got this script so far but i keep getting the following error message which i cannot figure out..

 

could any1 help..

 

error:Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\checklogin.php on line 18

Wrong Username or Password

 

code:

<?

 

$hostname_mysql_connect = "localhost";

$database_mysql_connect = "website_members";

$username_mysql_connect = "root";

$password_mysql_connect = "******";

$tblname_mysql_connect ="members";

 

$mysql_connect = mysql_pconnect($hostname_mysql_connect, $username_mysql_connect, $password_mysql_connect, $tblname_mysql_connect) or trigger_error(mysql_error(),E_USER_ERROR);

mysql_select_db($database_mysql_connect) or die(mysql_error());

 

$myusername = $_POST['myusername'];

$mypassword = $_POST['mypassword'];

 

$sql = "SELECT * FROM $tblname_mysql_connect WHERE username= '$myusername' and password='$mypassword'";

$result = mysql_query($sql);

 

$count = mysql_num_rows($result); (line 18)

if($count==1) {

session_register("myusername");

session_register("mypassword");

header("location:login_success.php");

}

else {

echo "Wrong Username or Password";

}

 

?>

 

 

many thanx

 

 

[timmah1]

try this instead of that tutorial

 

your form

<form action="<?php echo $SCRIPT_NAME; ?>" method="POST">
<table>
<tr>
    	<td>Email:</td>
      <td><input type="textbox" name="email">
    </tr>
    <tr>
    	<td>Password:</td>
        <td><input type="password" name="password">
    </tr>
    <tr>
    	<td></td>
        <td><input type="submit" name="login" value="Log In">
    </tr>
</table>
</form>

 

then the code to process the form

$dbhost = "localhost";
$dbuser = "USERNAME";
$dbpassword = "PASSWORD";
$dbdatabase = "DATABASE;

$db = mysql_connect($dbhost, $dbuser, $dbpassword);
mysql_select_db($dbdatabase, $db);

$config_sitename = "YOURSITENAME";
$config_basedir = "PATH TO YOUR SITE";

$loginsql = "SELECT * FROM members WHERE username= '" . $_POST['username'] . "' AND password = '" . $_POST['password'] . "'";
$loginres = mysql_query($loginsql);
$numrows = mysql_num_rows($loginres);

if($numrows == 1)
{
	$loginrow = mysql_fetch_assoc($loginres);

		session_register("SESS_LOGGEDIN");
		session_register("SESS_USER");
		session_register("SESS_COMPANY");
		session_register("SESS_USERID");

		$_SESSION['SESS_LOGGEDIN'] = 1;
		$_SESSION['SESS_USER'] = $loginrow['username'];
		$_SESSION['SESS_COMPANY'] = $loginrow['company'];
		$_SESSION['SESS_USERID'] = $loginrow['user_id'];

		header("Location: " . $config_basedir);
}
else 
{
		header("Location: http://" . $HTTP_HOST . $SCRIPT_NAME . "?error=incorrect");
}
switch($_GET['error']) {

		case "incorrect":
			echo "<font color=#FF0000><strong>Incorrect email/password</strong></font>";
		break;

	}

[PFMaBiSmAd]

The error means the mysql_query() failed. Since there is no error checking of the result of the mysql_query(), the code blindly continues execution and attempts to access a non-existent result resource.

 

The original code has error checking on the pconnect and select database functions but it has no error checking on the mysql_query(). The examples in the php manual under mysql_query() show basic error checking that will get your code to tell you why the query failed.

 

@timmah1, the code you posted has absolutely no error checking in it and it will fail for same reason as the original code.

[ma5ect]

thanx 4replies. But i dont understand da debuggin process. I am pretty new to php code

[.josh]

http://www.phpfreaks.com/tutorial/debugging-a-beginners-guide

[ma5ect]

http://www.phpfreaks.com/tutorial/debugging-a-beginners-guide

 

Hi,

 

i changed the code in the query to :

$sql = "SELECT * FROM $tblname_mysql_connect WHERE username= '$myusername' and password='$mypassword'";

$result = mysql_query($sql) or die(mysql_error());

 

and this message is shown:

 

Unknown column 'password' in 'where clause'

[waynew]

Make sure that you named your column password. I often find sometimes that I assumed I've called a table something, when in reality I used something like pass.

[ma5ect]

Make sure that you named your column password. I often find sometimes that I assumed I've called a table something, when in reality I used something like pass.

 

I just figured it out...i mis spelt the password field..

 

many thanx for all posts..