PHP is not returning me a number type
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by Tristan
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Published on 2010-04-24T02:09:45Z
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2010/04/24
2:13 UTC
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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
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