PHP is not returning me a number type
- by Tristan
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
