errm whats wrong with my mysql_fetch_row()

[buddysal]

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/buddysal/public_html/login.php on line 11

$_GET['$username'];
$_GET['$password'];

$user = "buddysal_admin";
$pass = "XXXXXX";
$db = "buddysal_USERS";
$link = mysql_connect( "localhost", $user, $pass );
mysql_select_db( $db, $link );

$check = mysql_fetch_row("SELECT * FROM MEMBERS WHERE username=['$username']");

if (! $password = $check['1'] ) {
print "Login Succeeded, Redirecting to members area!";
} else {
print "Login Failed, Singup for a free account first!";
}

[JDcrack]

Try this solution:

//Change:
$_GET['$username'];
$_GET['$password'];

//To:
$username = $_GET['$username'];
$password = $_GET['$password'];

//Change:
check = mysql_fetch_row("SELECT * FROM MEMBERS WHERE username=['$username']");

//To:
check = mysql_fetch_row("SELECT * FROM MEMBERS WHERE username='$username'");

Also for debugging purposes you may want to add an "or die" to the end of your mysql statement like:

mysql_fetch_row("SELECT * FROM MEMBERS WHERE username=['$username']")
	or die('<br>Query failed: ' . mysql_error());
//This will echo out a more exact reason for the query failure.

[buddysal]

ok i think im doing something wrong? i changed everything you said.. and i still get that mysql_fetch_row() error, it it says Query Failed: and dont give the reason...

$username = $_GET['$username'];
$password = $_GET['$password'];

$user = "buddysal_admin";
$pass = "XXXXXX";
$db = "buddysal_USERS";
$link = mysql_connect( "localhost", $user, $pass );
mysql_select_db( $db, $link );

$check = mysql_fetch_row("SELECT * FROM MEMBERS WHERE username='$username'")
    or die('<br>Query failed: ' . mysql_error());


if (! $password = $check['1'] ) {
print "Login Succeeded, Redirecting to members area!";
} else {
print "Login Failed, Singup for a free account first!";
}

[JDcrack]

is it possible that you are not even connecting to the DB?
try putting an or die after your DB connection

$link = mysql_connect( "localhost", $user, $pass )or die("mysql_connect Failed". mysql_error()); 
mysql_select_db( $db, $link ) ("mysql_select_db Failed". mysql_error()); 

PS
you will also want to edit your above POSTS and remove any username/password info, may be replace with $pass = XXXX

[JDcrack]

OOps I just caught your problem...
(I must be having a senior moment 🙂 )
You have not declared a mysql_query
Your select statement is your mysql_query, and then you can fetch_row from your mysql_query.
That is a vital step:

$result = mysql_query("SELECT * FROM clients WHERE name='$username'")
or die('<br>Query failed: ' . mysql_error());
//putting your SELECT into a mysql_query()

$check = mysql_fetch_row($result);
// fetching info from your mysql_query();

http://www.php.net/manual/en/function.mysql-query.php

[buddysal]

oh ok.. so it would be $query = mysql_query()
$check = mysql_fetch_rows($query);

[buddysal]

stupid crap.... now everything works but i always get a successful result even if the user dont exist..

[JDcrack]

Try using mysql_num_rows()

//Instead of:
if (! $password = $check['1'] ) {

//try
if (mysql_num_rows($query) > 0 && $password == $check[1] ) {
//if query returns at least 1 result and the passwords match
 print "Login Succeeded, Redirecting to members area!";
} else {
print "Login Failed, Singup for a free account first!";
}

Your original if statement was slightly confusing:
if (!$password
if the variable $password is not set
then you are checking if a not-set variable is equal to something.
this is illogical.

[buddysal]

well i added what u posted and it just flipped my result... now i always get returned false, even if the account does exist... im about to feel like quitting php lol its too confusing

[JDcrack]

trying changing $check[1]
to $check["password"]

This is assuming that "password" is the column name in your MySQL table.

[buddysal]

well ive totally screwed something up... so im just going to re-script the login... on the if(mysql_num_rows($query) bleh bleh... if i have it like that it returns false all the time if i put a ! in from of mysql_num_rows() it returns true all the time... for some reason its not checking the password field and if match = pass not = fail, and i did a test to see if it was checking the db so i did a print $row['1'] and it returned my user password.. so i dunno whats wrong with it..

[JDcrack]

Why not do it this way:

$query = mysql_query("SELECT FROM table WHERE username='$username' & password='$password');

if (mysql_num_rows($query) < 1) {
echo "not logged in";
}else{
echo "Login Correct!";
}

Also on another note,
Are you using the $_GET method for username and password for debugging purposes?
If not, you will want to use a different method.

[buddysal]

im using both $_GET[] to get the username from the submitted form... i have a quick login on my index page and i have login.php as the processor...

[JDcrack]

You are going to want to change the method to POST
and use $_POST in place of $GET

otherwise the username and password will apear in the URL

[buddysal]

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/buddysal/public_html/login.php on line 13
not logged in

thats what i get with the above

[JDcrack]

Did you make sure to change the action on your form to
action="POST" ?

[diego25]

Use "AND" instead of "&"

Diego

[buddysal]

yes, did both... and also am still having the same problems...

[JDcrack]

Please paste your new source here, so I can take a look at it.
Thanks

[buddysal]

here is my form script on index.php

<form action="login.php" method="POST">
<p><font size="2">Enter Username</font><br>
<input type="text" size="15"
name="username"
style="background-color: #DFDFDF"
style="border: thin solid #CDCDCD"
style="color: #626262"
style="letter-spacing: -1px"><br>
<font size="2">Enter Password</font><br>
<input type="password" size="15"
name="password"
style="background-color: #DFDFDF"
style="border: thin solid #CDCDCD"
style="color: #626262"
style="letter-spacing: -1px"><br>
<input type="submit" name="B1"
value="Login Now"
style="background-color: #DFDFDF"
style="border: thin solid #CDCDCD"><p>
</form>

here is my processor script as of right now on login.php

<?PHP
$username = $_POST['$username'];
$password = $_POST['$password'];

$user = "buddysal_admin";
$pass = "3585835";
$db = "buddysal_USERS";
$link = mysql_connect( "localhost", $user, $pass );
mysql_select_db( $db, $link );

$query = mysql_query("SELECT FROM MEMBERS WHERE username=['$username'] AND password=['$password']"); 

if (mysql_num_rows($query) < 1) { 
echo "not logged in"; 
}else{ 
echo "Login Correct!"; 
} 
?>

and heres the error i get with everything just like it is...

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/buddysal/public_html/login.php on line 13
not logged in