Изкарване от дб

kenwood10

Registered
Здравейте на всички това е кодът ми които ми изкарва от дб

PHP:
$query = "SELECT firm,SUM(win) as obshto,SUM(lost) as obshtozagubi,COUNT(firm) as kolko FROM users WHERE firm != '' GROUP BY firm ORDER BY SUM(win) DESC LIMIT $redove, $broinastranica";

Как да му кажа на този код да изкарва само firm където полето 'read' е = 0 Благодаря ви предварително.Дано сте ме разбрали
 
PHP:
$query = "SELECT firm,SUM(win) as obshto,SUM(lost) as obshtozagubi,COUNT(firm) as kolko FROM users WHERE firm != '' GROUP BY firm ORDER BY SUM(win) DESC LIMIT $redove, $broinastranica" and read='0' ;
 
И така пробвах не стана... ето я цялата заявка...

PHP:
$query = "SELECT firm,SUM(win) as obshto,SUM(lost) as obshtozagubi,COUNT(firm) as kolko FROM users WHERE firm != '' AND read='0' GROUP BY firm ORDER BY SUM(win) DESC LIMIT $redove, $broinastranica"  ;
$result = mysql_query($query) or die('Error, query failed');

:(
 
А каква е точно грешката?
PHP:
$query = "SELECT firm,SUM(win) as obshto,SUM(lost) as obshtozagubi,COUNT(firm) as kolko FROM users WHERE firm != '' AND read='0' GROUP BY firm ORDER BY SUM(win) DESC LIMIT $redove, $broinastranica"  ;
$result = mysql_query($query) or die([color=red]mysql_error()[/color]);
 
PHP:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'read='0' GROUP BY firm ORDER BY SUM(win) DESC LIMIT 0, 25' at line 1
 
PHP:
$query = "SELECT firm,SUM(win) as obshto,SUM(lost) as obshtozagubi,COUNT(firm) as kolko FROM users WHERE firm != '' GROUP BY firm ORDER BY SUM(win) and read='0' DESC LIMIT $redove, $broinastranica " ;
 
dakata__92 каза:
PHP:
$query = "SELECT firm,SUM(win) as obshto,SUM(lost) as obshtozagubi,COUNT(firm) as kolko FROM users WHERE firm != '' GROUP BY firm ORDER BY SUM(win) and read='0' DESC LIMIT $redove, $broinastranica " ;

+1
 

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