[darkcarnival]

hi,

i have this rating system that i got to set-up good but when i load the rate file it displays this error:

Rating: LINE 74:You have an error in your SQL syntax near '' at line 1

now i dont know what this means so could someone plz tell me what it is and how to fix it.

[Ryan]

Could you post the SQL command you tried to use, please?

[darkcarnival]

i dont know what it is it did it once i ran the script if u want the script reply and ill give it to u. now i checked the line and it is a insert into command. thats all i know.

[Ryan]

Could you post the code of the script then? The error you gave is too general for someone to give you a solution to.

[darkcarnival]

so u want the line that made the error or the whole script.

[Ryan]

Well, let's see that line first.

[darkcarnival]

ok heres that line:

$query = "SELECT COUNT(*) as num, ((COUNT(*)/(COUNT(*)+".$GLOBALS['m']."))*AVG(rate)+(".$GLOBALS['m']."/(COUNT(*)+".$GLOBALS['m']."))*".$GLOBALS['C'].") AS rate FROM ".$GLOBALS['table']." WHERE rate>0 AND rate<11 AND article=".$id;
$res = mysql_query($query) or die("<b>LINE 74</b>:".mysql_error());

[Ryan]

Ok you can give this a shot, I took out a lot of the parenthesis (I never use them) but I probably forgot something, somewhere.

PHP Code:

<?php

$query = "SELECT COUNT(*) AS num, COUNT(*) / COUNT(*) + {$GLOBALS['m']} * AVG(rate) + {$GLOBALS['m']} /(COUNT(*) + {$GLOBALS['m']} * {$GLOBALS['C']} AS rate FROM {$GLOBALS['table']} WHERE rate > 0 AND rate < 11 AND article= {'$id'}";



$res = mysql_query($query) or die("<b>LINE 74</b>:".mysql_error());

?>

[darkcarnival]

so copy and paste this where the old line was.

[Ryan]

Yes, try that and let me know how it goes.