Object of class mysqli_result could not be converted to string in

[kimla]

Hi.

 

I'm getting this error (see subject), and can't seem to find out why...

 

					
if($this->db->query("INSERT INTO blog VALUES('', '". $userid ."', '". $name ."', NOW()')") != true)
{
	$this->setError("Kunne ikke opprette blogg");
	return false;
}
else
{
	return true;
}

 

$this->db->query is simply:

public function query($var)
{
return $this->link->query($var); 
}

 

in my db-class. Yes, I know it might be a messy way to set it up, but still, should'nt it work?

 

All answers appreciated.

[kimla]

Btw, in a bit of a hurry here... Schoolwork.. =)

[teng84]

school work?

I guess it bad idea to do your home work

[ToonMariner]

It looks like you are returning the resource id...

 

use the id and fetch_array_assoc or similar to get the actual results

 

[aschk]

$this->db is what? a reference to itself?

 

Because your query function is part of your class, you want to use

 

$this->query(...)