PHP is not returning me a number type

Question

Hello, i tryed to follow that great tutorial (STAR rating with css : http://stackoverflow.com/questions/1987524/turn-a-number-into-star-rating-display-using-jquery-and-css)

but i've just a big problem : When i do

<span class="stars">1.75</span>

or

$foo='1.75';
echo '<span class="stars">'.$foo.'</span>

the stars is correctly shown, but as soon as i do :

while($val = mysql_fetch_array($result)) 
{  
$average =  ($val['services'] + $val['serviceCli']  + $val['interface'] + $val['qualite'] + $val['rapport'] )  / 5 ;

<span class="stars">.$average.</span>
}

the stars stops working

i double checked the data type in mysql : they're all TINYINT(2)

and i tryed that :

$average = intval($average);

but it's still not working,

Thank you

Answer 1

no, var_dump($val) gives me ALL the value i get from my big request (text, varchar, int) [example :

array(38) { [0]=>  string(28) "http://www.crystal-serv.com/" 
["siteweb"]=>  string(28) "http://www.crystal-serv.com/" 
[1]=>  string(1) "0" ["offreDedie"]=>  string(100) 
"tick rouge" [2]=>  string(1) "0" ["coupon"]=> 

..........

Answer 2

I think your problem may be that the value you have is greater than the 5 allowed in that example.

What you want to do is weight the items such that the total for $average is less than or equal to 5.

$average = (
    ( $val['services'] / $maxServices )
    + ( $val['serviceCli'] / $maxServiceCli )
    + ( $val['interface'] / $maxInterface )
    + ( $val['qualite'] / $maxQualite )
    + ( $val['rapport'] / $maxRapport )
);

The weighting could be even, so each of the values will be less than or equal to 1, or they could have different weights so services is worth more than qualite (and so on).