[SOLVED] mysql_fetch_array problems.

[Noxxia]

I've spent the last hour and a half trying to figure out what I'm doing wrong by looking through various forums without any luck.

 

Basically, I'm getting this wonderful error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

 

I originally had a string with a bunch of variables in it and thought that was the problem.  So I just hard coded a MySQL query that works in the Starfields, Inc. SQL tab.

 

SELECT `Noxxia9`.`UserData`.* FROM `Users` INNER JOIN `UserData` ON `Users`.`UserId` = `UserData`.`UserId` AND `Users`.`Username` = 'flipsIX'

 

This produces the proper result I'm looking for, So I then put that into my php function, knowing that obviously without variables it won't function correctly, but at least I shouldn't get the error because I won't be returning an empty result.

 

$queryString = "SELECT `Noxxia9`.`UserData`.* FROM `Users` INNER JOIN `UserData` ON `Users`.`UserId` = `UserData`.`UserId` AND `Users`.`Username` = 'flipsIX'";
$selectQuery = mysql_query($queryString, $conDB);
return mysql_fetch_array($selectQuery);

 

I will still get the error.  I'm new to SQL and have only been working with it for a few days.  My php isn't great, but i'm not a complete idiot about it.  There's obviously something I'm not catching though.  Anyone have any ideas?

 

Btw, any of my other mysql_query function calls with a similar sort of $queryString works just fine.  This is the only one to work and the only one with a JOIN, but I coudln't possibly be the join could it? I don't understand how that would matter since mysql_query returns a resource identifier, but maybe there's soemthing about that I don't understand :/

 

Thanks

[.josh]

you said you have another mysql_query that works.  Is it in the same script as this one?  I'm wondering if you failing to connect to the db in the first place. 

[.josh]

 

I see a return in there.  That means you're in a method or function.  And you are using $conDB.  Is this set outside of the function?  You do not need to assign mysql_connect to a variable.  But if you do, it is not automatically (super)global.  You can probably remove it from mysql_query in your function and php will try to use it automatically.  If that don't work, put

 

global $conDB;

 

inside your function.

[Noxxia]

Yeah, on the same page in fact.  This is my attempt at getting a users data from a seperate table after authenticating for login.  I'm able to compare the username and password just fine, then when it gets to that stuff, which isi n a seperate function, that won't work.  My actual code looks like this:

function getUserData($username) {
	$queryString = "SELECT `Noxxia9`.`UserData`.* FROM `Users` INNER JOIN `UserData` ON `Users`.`UserId` = `UserData`.`UserId` AND `Users`.`Username` = '$username'";
	//$queryString = "SELECT `Noxxia9`.`UserData`.* FROM `Users` INNER JOIN `UserData` ON `Users`.`UserId` = `UserData`.`UserId` AND `Users`.`Username` = 'flipsIX'";
	$selectQuery = mysql_query($queryString, $conDB);
	$result = mysql_fetch_array($selectQuery);
	return $result;
}

 

here's anotherone that works on the same page.

 

function pwCheck($username, $password, $conDB, $dbName, $tableName) {
	$queryString = "SELECT * FROM `$dbName`.`$tableName` WHERE `Username` = '$username' AND `Password` = '$password'";
	$selectQuery = mysql_query($queryString, $conDB);
	$result = mysql_fetch_array($selectQuery);
	if ($result == "") {
		return false;
	} else {
		return true;
	}
}

 

 

these used to be on the same page of generic functions i've made for myself in case I need them later, but I moved the first one listed (the one i'm having trouble with) to the same page as the login page thinking that maybe I messed up passing parameters.  I turned it from generic into very literal to help me narrow down the problem, and actually generated another, but this might help solve it. 

 

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource <dir stuff> on line 54

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource <dir stuff> on line 55

 

line 54 and line 55 are in getUserData().

 

the second function listed, works just fine and is only called 2 lines prior to the first function listed.

 

$password = encryptPW();
$validUser = pwCheck($_POST["username"], $password, $conNoxxia9, "Noxxia9", "Users");
if ($validUser) {
$userData = getUserData($_POST["username"]);

[Noxxia]

 

I see a return in there.  That means you're in a method or function.  And you are using $conDB.  Is this set outside of the function?  You do not need to assign mysql_connect to a variable.  But if you do, it is not automatically (super)global.  You can probably remove it from mysql_query in your function and php will try to use it automatically.  If that don't work, put

 

global $conDB;

 

inside your function.

 

ooops...you made me realize one mistake.  I forgot to do that after moving the function.

 

i still end up with my original error rather than the 2 errors now.  that was the reason I got two of them.

[.josh]

..and i think that's what your problem is. In the one you listed that works, you are passing it as an argument.

[Noxxia]

Still not working.  Do I maybe need to close the connection and then reopen it?

[Noxxia]

bad sql command.  forgot to put USE >.<