Question

php多条件查询问题

$num11 = DB::result_first('SELECT COUNT(*) FROM '.DB::table('quote_data')." WHERE variety=$fid AND uality=$i+1 AND view=1 ");
$num12= DB::result_first('SELECT COUNT(*) FROM '.DB::table('quote_data')." WHERE variety=$fid AND uality=$i+1 AND view=2 ");
$num13= DB::result_first('SELECT COUNT(*) FROM '.DB::table('quote_data')." WHERE variety=$fid AND uality=$i+1 AND view=3 "); 以上代码帮忙改下，怎样才能使代码更简单，让它一次完成，不再查询3次，我的目的就是要取得相同表里，3个不同值的数据的数量有人告诉一下方法：$num= DB::result_first('SELECT COUNT(*) FROM '.DB::table('quote_data')." WHERE variety=$fid AND uality=$i+1 AND view in (1,2,3) "); 但是这样的话，怎样得知3个$num，每一个的数量是多少呢？

Answer 1

$sql = 'SELECT view, COUNT(*) FROM '.DB::table('quote_data')." WHERE variety=$fid AND uality=$i+1 AND view in (1,2,3)';while($rs = mysql_fetch_array( mysql_query($sql))){switch($rs['view']){case 1: $view1 = $rs ['num']; break;case 2: $view2 = $rs ['num']; break;.................}}你可以把query出来的数组打印出来看看啊,找找共同点或者这样while($rs = mysql_fetch_array( mysql_query($sql))){ $views [ $rs ['view'] ] = $rs['num'];}</p>

Answer 2

SELECT view,COUNT(view) FROM '.DB::table('quote_data')." WHERE variety=$fid AND uality=$i+1 AND (view=1 || view=2 || view=3) group by view ");

Answer 3

楼主在玩Discuz! 插件开发吧，呵呵。