Warning: mysql_query(): supplied argument is not a valid MySQL-Link

[tecmeister]

Hi everyone,

 

I'm getting this error and I don't know why.

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\wamp\www\users\include\database.php on line 16

 

I am proberly missing something.

 

Here is the code that I'm referring it from

$first = $_POST['first'];
$last = $_POST['last'];
$email = $_POST['email'];
$user = $_POST['user'];
$pass = $_POST['pass'];

$sql = "INSERT INTO users (firstname, lastname, email, username, password)";
$sql .= " VALUES ('$first', '$last', '$email', '$user', '$pass')";
$result = $database->insert_user($sql);

 

This is what I'm sending it to:

function query($sql){
	$result = mysql_query($sql,$this->connection);
	$this->confirm_query($result);
	return $result;
}

private function confirm_query($result){
	if(!$result){
		die("Database query failed: " . mysql_error());
	}
}

 

Thanks for your help,

 

tecmeister.

[Maq]

This is what I'm sending it to:

 

No it's not, you're sending it to 'insert_user'.

[tecmeister]

Sorry, that was beacuse I pressed ctrl + z before I copied it.

When I send it to query I get the same problem.

[Maq]

Add this:

 

      $result = mysql_query($sql,$this->connection) or die(mysql_error());

[tecmeister]

I had that originally, but I change it to the tutorial way and I still have the same warning.

Is there something wrong with the $sql variable?

[Maq]

I had that originally, but I change it to the tutorial way and I still have the same warning.

Is there something wrong with the $sql variable?

 

Well what's the warning say?  The SQL syntax looks valid to me...  Maybe you should try to echo out '$sql' to see the string that's actually being passed to your function.

[tecmeister]

I have also tried that and print.

This is what I get.

INSERT INTO users (firstname, lastname, email, username, password) VALUES ('Johnny', 'McCaffery', 'notanotherperson@hotmail.com', 'tecmeister', 'password').

 

Can you see and error?

 

Thanks for your help, much appreciated

 

temeister.

[Maq]

Looks fine to me.

 

You said earlier that you got a warning?  Please share.  :'(

[tecmeister]

Here is the warning:

 

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\wamp\www\users\include\database.php on line 16

 

That line is this:

$result = mysql_query($sql,$this->connection) or die(mysql_error());

 

[Maq]

Oh jeez...

 

Isn't connection supposed to be a variable?

[lonewolf217]

it should be an optional parameter right?

 

What happens when you take out $this->connection entirely

[tecmeister]

When I take away $this->connection.

It cant connect to the database.

 

Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\wamp\www\users\include\database.php on line 17

 

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\wamp\www\users\include\database.php on line 17

Access denied for user 'ODBC'@'localhost' (using password: NO)

[Maq]

Please post your entire database class.

 

In your constructor you have your connection string similar to?

 

$this->connection = mysql_connect(DB_SERVER, DB_USER, DB_PASS) or die(mysql_error());

[tecmeister]

<?php
class mysqldatabase{

private $connection;

function __content(){
	$this->open_database;
}

function open_database(){
	$this->connection = mysql_connect("localhost","teci","********")or die("Connect Error".mysql_error());
	$this->connection = mysql_select_db("users")or die("Select Database Error".mysql_error());
}

function query($sql){
	$result = mysql_query($sql,$this->connection) or die(mysql_error());
	return $result;
}


}
$database = new mysqldatabase;
return $database;
?>

[tecmeister]

I think that I have found the error

$this->open_database();

 

Sorry still got that warning.

[Maq]

Where is the private function confirm_query()?  You need to post consistent code.

 

Try using this class:

 

class mysqldatabase{

   private $connection;
   
   /* Class constructor */
   function __construct(){
      $this->connection = mysql_connect("localhost","teci","********")or die("Connect Error".mysql_error());
      $this->connection = mysql_select_db("users")or die("Select Database Error".mysql_error());
   }
   
   function query($sql){
      $result = mysql_query($sql,$this->connection) or die(mysql_error());
      return $result;
   }
   
}

$database = new mysqldatabase;
return $database;
?>

 

[Ken2k7]

Maq - typo in your last line in the constructor. Although, I'm not sure where you're returning $database to either.

[anupamsaha]

Try this:

 

$sql = "INSERT INTO `users` (`firstname`, `lastname`, `email`, `username`, `password`)";

 

I think there is a conflict between the field password and MySQL function PASSWORD. So, its better to enclose all the MySQL table name and field names with backquote (`) character in order to avoid the conflict.

 

Hope this will help.