Тема: помогите написать (WAP/WEB мастеру/PHP/MySQL)

Engine (16 Oct 2013 в 20:41)
$user_m = mysql_result(mysql_query(\"SELECT COUNT(*) FROM `user` WHERE `pol` = \'1\'\"),0); 
$user_w = mysql_result(mysql_query(\"SELECT COUNT(*) FROM `user` WHERE `pol` = \'0\'\"),0); 
echo \'(\'.$user_m.\'|\'.$user_w.\')\'; 
// Как то так

Кредитор (16 Oct 2013 в 20:45)
хз. пробуй так:
$count = mysql_fetch_assoc(querys(\"SELECT COUNT(CASE WHEN pol = 1 THEN 1 ELSE NULL END) AS p, COUNT(CASE WHEN pol < 1 THEN 1 ELSE NULL END) AS d FROM user\"));
echo \'Парни: \' . $count[\'p\'] . \' Девченки \' . $count[\'d\'];

Источник: http://www.dcms-social.ru/forum/20/41/9494/