Question

Here is my error:

Warning: mysql_query() expects parameter 2 to be resource, null given…

This refers to line 23 of my code which is:

$result = mysql_query($sql, $connection)

My entire query code looks like this:

$query = "SELECT * from users WHERE userid='".intval( $_SESSION['SESS_USERID'] )."'"; 
                $result = mysql_query($query, $connection)
                or die ("Couldn't perform query $query <br />".mysql_error());

                $row = mysql_fetch_array($result);

I don’t have a clue what has happpened here. All I wanted to do was to have the value of the users ‘fullname’ displayed in the header section of my web page. So I am outputting this code immediately after to try and achieve this:

echo 'Hello '; echo $row['fullname'];

Before this change, I had it working perfectly, where the session variable of fullname was echoed $_SESSION[‘SESS_NAME’]. However, because my user can update their information (including their name), I wanted the name displayed in the header to be updated accordingly, and not displaying the session value.

Answer 1

Your $connection variable is NULL that’s what your error message is referring to.

Reason being is that you have not called mysql_connect. Once called it will assign you a resource where you can set it to the $connection variable, thus being non-null.

As an example:

$connection = mysql_connect('localhost', 'mysql_user', 'mysql_password');
// now $connection has a resource that you can pass to mysql_query
$query = "SELECT * from users WHERE userid='".
                                         intval( $_SESSION['SESS_USERID'] )."'"; 
$result = mysql_query($query, $connection)