[Resolved] OOP Hates Me. DB: NO Pass passed.

[Kudose]

Hello again all. I apologize before hand, but this script is really ticking me off.

First, I cannot get the varables to pass to my classes connection. It doesnt recognize the $dbUser or $dbPass or any of that, so I get connection errors.

Secondly, when it did connect, once upon a time, it never reports data or errors. Only the static content.

Any help would be great!

<?php
class dbConnection {
	var $dbLoc;
	var $dbUser;
	var $dbPass;
	var $dbName;
	var $items = array();
	var $result;
	var $rows;
	var $connection;
	var $connect;
	var $death;

function dbConnection($dbLoc, $dbUser, $dbPass, $dbName){
	if($this->connect = mysql_connect($this->dbLoc, $this->dbUser, $this->dbPass)){
		if($this->connection = mysql_select_db($this->dbName, $this->connection)){
			return $this->connection;
		} else {
			return mysql_error();
		}
	} else {
		return mysql_error();
	}
}

function dbQuery($sql){
	if(mysql_query($sql, $this->connection)){
		return true;
	} else {
		return mysql_error();
	}
}

function dbFetch($sql){
$this->items = array();
	if(mysql_query($sql, $this->connection)){
		$item = array();
		while($item = mysql_fetch_array($sql)){
			$this->items[] = $item;
		}
		if(!mysql_error()){
			return $this->items;
		} else {
			return mysql_error();
		}
	} else {
		return mysql_error();
	}
}

function dbRows($sql){
	$this->result = mysql_query($sql, $this->connection);
	$this->rows = mysql_num_rows($this->result);

	if(!mysql_error()){
		return $this->rows;
	} else {
		return mysql_error();
	}
}

function dbKill(){
	$this->death = mysql_close($this->connection);
	if($this->death){
		return true;
	} else {
		return mysql_error();
	}
}
}

$myDb = new dbConnection("localhost", "xxx_xxx", "xxx", "xxx_xxx");

$result = $myDb->dbFetch("SELECT * FROM config");

echo "The result for title was: ".$result[0]['title'];
?>

[Kudose]

Oh yes, I forgot to include that I am hosted on PHP 4.3.9.

TIA.

[Kudose]

I have determined that the problem lies within passing the variables into the class. If I set the varaiables from within the class than I get a connection to MySQL.

Now....I dunno what's wrong.

[Kudose]

Ok then...

I have fixed the connection issue...

Now, when I try to use this function, I get resource link errors.

function dbFetch($sql){
		$this->items = array();
		$this->result = mysql_query($sql, $this->connection);
		$item = array();
		while($item = mysql_fetch_array($sql)){
			$this->items[] = $item;
		}
		if(!mysql_error()){
			return $this->items;
		} else {
			return mysql_error();
		}
	}

If I run an if else on $this->connection from within the function, it states that the connection exists.

What would cause

Connection is still alive!
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in...

?

TIA.

[tsinka]

Hi,

the problem is that you first need to set class variables in the constructor. Something like this (simple):

function dbConnection($dbLoc, $dbUser, $dbPass, $dbName) {
    $this->dbLoc  = $dbLoc;
    $this->dbUser = $dbUser;
    $this->dbPass = $dbPass;
    $this->dbName = $dbName;
    if($this->connection = mysql_connect($this->dbLoc, $this->dbUser, $this->dbPass)){
        if(mysql_select_db($this->dbName, $this->connection)){
            return $this->connection;
        } else {
            return mysql_error();
        }
    } else {
        return mysql_error();
    }
} 

I'd suggest to check out some of the classes available at e.g. the code library of PHPBuilder.

I think the class needs some tweaking.

Thomas

[Kudose]

Thanks for you reply Thomas.

I did figure that one out after looking at a few more examples of classes on the net.

Now I have an issue of my connection dying or something.

I call out to make sure that $this->connection is still available and working, it states that it is. But I get MySQL resource link errors on both the query and the fetch_array...

Any ideas?

Thanks thus far!

<?php
class dbConnection {
	var $items = array();
	var $result;
	var $rows;
	var $connection;
	var $connect;
	var $death;

function dbConnect(){
	if(mysql_connect($this->dbLoc, $this->dbUser, $this->dbPass)){
	$this->connect = mysql_connect($this->dbLoc, $this->dbUser, $this->dbPass);
		if(mysql_select_db($this->dbName, $this->connect)){
			$this->connection = mysql_select_db($this->dbName, $this->connect);
			return $this->connection;
		} else {
			return mysql_error();
		}
	} else {
		return mysql_error();
	}
}

function dbQuery($sql){
	$this->sql = $sql;
	if(mysql_query($this->sql)){
		return true;
	} else {
		return mysql_error();
	}
}

function dbFetch($sql){
	$this->sql = $sql;
	$this->items = array();
	$this->result = mysql_query($this->sql, $this->connection);
	$item = array();
	while($item = mysql_fetch_array($this->sql, $this->connection)){
		$this->items[] = $item;
	}
	if(!mysql_error()){
		return $this->items;
	} else {
		return mysql_error();
	}
}
}
$myDb = new dbConnection();

$myDb->dbLoc = "localhost";
$myDb->dbUser = "xxx_xxx";
$myDb->dbPass = "xxx";
$myDb->dbName = "xxx_xxx";

$myDb->dbConnect();

$result = $myDb->dbFetch("SELECT * FROM config");

echo "The result for title was: ".$result[0]['title'];
?>

[PHPMagician]

You're connection is not valid. You set $this->connection to the return of mysql_select_db when all this function do is return true or false, not a resource. You then use this in the second parameter of mysql_select in you dbFetch method. It really is an invalid resource link.

In your method dbConnect you have this:

function dbConnect(){
        if(mysql_connect($this->dbLoc, $this->dbUser, $this->dbPass)){
        $this->connect = mysql_connect($this->dbLoc, $this->dbUser, $this->dbPass);
            if(mysql_select_db($this->dbName, $this->connect)){
                $this->connection = mysql_select_db($this->dbName, $this->connect);
                return $this->connection;
            } else {
                return mysql_error();
            }
        } else {
            return mysql_error();
        }
    } 

Which I must say is badly designed. You are calling mysql_connect and mysql_select_db twice, which all slows down you're script. Think instead of doing something similar to:

function dbConnect()
{
    if (($this->connection = mysql_connect($this->dbLoc, $this->dbUser, $this->dbPass)) === false) {
        echo mysql_error();
        return false;
    }

if (mysql_select_db($this->dbName, $this->connection) === false) {
    echo mysql_error();
    return false;
}
}

That should solve you're problems.

Hope that helps

[PHPMagician]

It might also be a good idea to get into the habit of commenting you're code. This makes it easier for others to indentify problems for others and yourself. If you commented the reason why you set this->connection with the return value of mysql_select_db, you may have found the error yourself.

Good Luck

[Kudose]

Thank you. 🙂

[Kudose]

Would is be more efficient to execute MySQL queries withthe resource link as the second parameter, or without?

I ask becuase with the resource link as the second parameter, wouldn't it loop back through and reconnect everytime?

Or does it even matter?

Thanks. 🙂

[PHPMagician]

It will no redo the connection, but simply point to the resource. If you don't specify the connection, the last opened connection is assumed to be the one to use - this could cause problems if you used multiple connections. I wouldn' take it as fact, but I would say it is better perfomance wise to provide a direct reference to the connection resource rather than making the module doing all the work.