while($row1 = mysql_fetch_array($result1,MYSQL_ASSOC))
            {
            $values[] = $row1[TestScore];
            }

i think it returns nothing, as i get an error message:-

Warning: Division by zero in /home/fhlinux196/a/askmultimedia.co.uk/user/htdocs/FW/website/example1.php on line 30

line 30 is:-

$column_width = $width / $columns ;

which means that $columns has nothing in it! which is set by line 17:-

$columns = count($values);

which in turn is set by the array, is there a way i could check if there is anything in the array?




My new web design domain
www.askmultimedia.co.uk

Add
print_r($values);
to check the value of $values.

Everything is temporary, some things are just more temporary than others... except for death, that seems to be pretty permanent

Everything looks ok from here. If $values is not returning anything I would put an echo $row1['TestScore']; in the while loop just to be sure the sql statment is returning what you expect. If it does not I would print_r($result1); to see what you are really getting back.

Everything is temporary, some things are just more temporary than others... except for death, that seems to be pretty permanent

while($row1 = mysql_fetch_array($result1,MYSQL_ASSOC))
            {
            $values[] = $row1[TestScore];
            }
        print_r($values);

That shows me nothing! so im guessing the array is blank, do i need to create the array?

Sorry if im messing with your brain!lol

thanks

My new web design domain
www.askmultimedia.co.uk

We posted at about the same time. Read my post above. "If $values is not returning anything I would put an echo $row1['TestScore']; in the while loop just to be sure the sql statment is returning what you expect. If it does not I would print_r($result1); to see what you are really getting back."

Everything is temporary, some things are just more temporary than others... except for death, that seems to be pretty permanent

also it is good practice to enclose TestScore with a single quote such as $row1['TestScore']; Without the single quote php first looks for a defined constant of TestScore.

Everything is temporary, some things are just more temporary than others... except for death, that seems to be pretty permanent

lol,

yeah its finally worked, it wasnt my code that was wrong, it was the querystring at the top of the code, which was blank, because i hadnt passed the value accross sucessfully!

Only by chance did i notice that! really been making me think i was a failure, in life in general!lol

Thanks for your help mate, its all fixed and working now!

cheers,
Ash

My new web design domain
www.askmultimedia.co.uk

Glad I could help. I once spent five hours to find out I had a dollar sign in the wrong place. Talk about feeling like a failure. Now if I could only fix my CSS problem that I have been working on for a week.

Everything is temporary, some things are just more temporary than others... except for death, that seems to be pretty permanent

hey, thanks, it works now, it wasnt the code that was the problem, it was my querystring $testID variable that was blank, in turn returning a blank SQL query!lol, fixed it now tho!

Oh, one more thing, i need one value in this array to be 100, without the DB is there a way in which i could do this?

while($row1 = mysql_fetch_array($result1,MYSQL_ASSOC))
            {
            $values[] = $row1[TestScore];
        $values[0] = 100;
            }

ive tried this but it overwrights the first value from the database!

cheers
Ash

My new web design domain
www.askmultimedia.co.uk

$i = 0;
             while($row1 = mysql_fetch_array($result1,MYSQL_ASSOC))
             {
            $i++;
            $values[$i] = $row1[TestScore];
            $values[0] = 100;
            }

there we go, that works, thanks again, and if i could help you with CSS i would, but trust me im terrible at that too!!

My new web design domain
www.askmultimedia.co.uk