【发布时间】:2011-06-28 00:00:07
【问题描述】:
$a = mysql_query("SELECT COUNT(*) as `count_1` FROM `table_1` WHERE `this_1` = '1'");
$b = mysql_fetch_assoc($a);
$c = mysql_query("SELECT COUNT(*) as `count_2` FROM `table_1` WHERE `this_2` = '1'");
$d = mysql_fetch_assoc($c);
【问题讨论】: