[SOLVED] "Supplied argument is not a valid MySQL result resource"

[Cory94bailly]

Error I'm getting:

 

<b>Warning</b>:  mysql_fetch_array() [<a href='function.mysql-fetch-array'>function.mysql-fetch-array</a>]: The result type should be either MYSQL_NUM, MYSQL_ASSOC or MYSQL_BOTH. in <b>/home/content/m/a/r/markbailly/html/fcs/includes/loggedin.php</b> on line <b>15</b><br />
<br />
<b>Warning</b>:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>/home/content/m/a/r/markbailly/html/fcs/includes/loggedin.php</b> on line <b>15</b><br />

 

 

Whole php code:

 

<?
if(isset($_COOKIE['ID_fcs_member']))
{
//Connect to Members DB
mysql_connect('***', '***', '***) or die(mysql_error());
mysql_select_db('***') or die(mysql_error());

//Look for the cookies
$username1 = $_COOKIE['ID_fcs_member'];
$pass1 = $_COOKIE['Key_fcs_member'];
$username2 = $_COOKIE['ID_fcs_team'];
$pass2 = $_COOKIE['Key_fcs_team'];
$check1 = mysql_query("SELECT * FROM members WHERE username = '$username1'")or die(mysql_error());
$check2 = mysql_query("SELECT * FROM team WHERE username = '$username2'")or die(mysql_error());
while(mysql_fetch_array($check1, $check2))
{
//If the cookie exists, show the "Logout" button.
{
?>
<a href="logout.php">Logout</a>
<?
}
}
}
else

//If the cookie does not exist, show the "Login" button.
{
?>
<a href="login.php">Login</a>
<img src="images/splitter.gif" class="splitter" alt="" />
<a href="register.php">Register</a>
<?
}
?>

 

 

Line 15:

 

while(mysql_fetch_array($check1, $check2))

[DarkWater]

You can't fetch two row sets in that manner.

[Cory94bailly]

You can't fetch two row sets in that manner.

 

Oh..

[Cory94bailly]

Explainations on how to do it would help

[BlueSkyIS]

if (mysql_numrows($check1) > 0) { // a record was found with the first query

} else if (mysql_numrows($check2) > 0) { // a record was found with the first query

}

 

personally, i'd run the first query, see if there are any records, then run the second query. no need to do both unless you need to do it that way.

[Cory94bailly]

if (mysql_numrows($check1) > 0) { // a record was found with the first query

} else if (mysql_numrows($check2) > 0) { // a record was found with the first query

}

 

personally, i'd run the first query, see if there are any records, then run the second query. no need to do both unless you need to do it that way.

 

I now get this:

 

<b>Parse error</b>:  syntax error, unexpected $end in <b>/home/content/m/a/r/markbailly/html/fcs/includes/loggedin.php</b> on line <b>38</b><br />

 

My full code:

 

<?
if(isset($_COOKIE['ID_fcs_member']))
{
//Connect to Members DB
mysql_connect('***', '***', '***') or die(mysql_error());
mysql_select_db('***') or die(mysql_error());

//Look for the cookies
$username1 = $_COOKIE['ID_fcs_member'];
$pass1 = $_COOKIE['Key_fcs_member'];
$username2 = $_COOKIE['ID_fcs_team'];
$pass2 = $_COOKIE['Key_fcs_team'];
$check1 = mysql_query("SELECT * FROM members WHERE username = '$username1'")or die(mysql_error());
$check2 = mysql_query("SELECT * FROM team WHERE username = '$username2'")or die(mysql_error());
if (mysql_numrows($check1) > 0) { // a record was found with the first query

} else if (mysql_numrows($check2) > 0) { // a record was found with the first query

{
//If the cookie exists, show the "Logout" button.
{
?>
<a href="logout.php">Logout</a>
<?
}
}
}
else

//If the cookie does not exist, show the "Login" button.
{
?>
<a href="login.php">Login</a>
<img src="images/splitter.gif" class="splitter" alt="" />
<a href="register.php">Register</a>
<?
}
?>

 

Line 38:

 

?>

[DarkWater]

You missed a }.

[Cory94bailly]

You missed a }.

 

Ok..

 

I just tried putting it at the end and the script showed nothing..

 

Where should I put it?

[DarkWater]

Wherever you want the IF statement to stop, lmfao.

[Cory94bailly]

Wherever you want the IF statement to stop, lmfao.

 

Well.. I'm not good at php so I didn't know..........

 

 

Here's my script now:

 

<?
if(isset($_COOKIE['ID_fcs_member']))
{
//Connect to Members DB
mysql_connect('p41mysql145.secureserver.net', 'cory_fcs', 'Sils1234') or die(mysql_error());
mysql_select_db('cory_fcs') or die(mysql_error());

//Look for the cookies
$username1 = $_COOKIE['ID_fcs_member'];
$pass1 = $_COOKIE['Key_fcs_member'];
$username2 = $_COOKIE['ID_fcs_team'];
$pass2 = $_COOKIE['Key_fcs_team'];
$check1 = mysql_query("SELECT * FROM members WHERE username = '$username1'")or die(mysql_error());
$check2 = mysql_query("SELECT * FROM team WHERE username = '$username2'")or die(mysql_error());
if (mysql_numrows($check1) > 0) { // a record was found with the first query

} else if (mysql_numrows($check2) > 0) { // a record was found with the first query

{
//If the cookie exists, show the "Logout" button.
{
?>
<a href="logout.php">Logout</a>
<?
}
}
}
}
else

//If the cookie does not exist, show the "Login" button.
{
?>
<a href="login.php">Login</a>
<img src="images/splitter.gif" class="splitter" alt="" />
<a href="register.php">Register</a>
<?
}
?>

 

I am getting no actual "errors"..

 

But when I am logged out, it works.. if you log in, it just shows nothing.

[Cory94bailly]

BUMP

[DarkWater]

1) Annoying bumps are annoying.

 

2) I think it's mysql_num_rows and not mysql_numrows. >_>

[Cory94bailly]

1) Annoying bumps are annoying.

 

2) I think it's mysql_num_rows and not mysql_numrows. >_>

 

1.) Sorry

 

2.) Ooh, nice fine.. I'll try it now.

[Cory94bailly]

Still blank if I log in.