Как сделать без .menu ради уменьшения количества SQL запросов?

@getStatHash[week][w]
	^if($week){
		$result[^hash::create[]]
		^week.menu{
			$w($week.n)
			$result.$w[^oSql.hash{
				SELECT
					app_id
					,COUNT(*) AS cnt
				FROM
					weekly
				WHERE
					dt > '$week.from'
				AND
					dt < '$week.to'
				GROUP BY
					app_id
			}]
			$result.$w.total(^oSql.int{
				SELECT
					COUNT(*)
				FROM
					weekly
				WHERE
					dt > '$week.from'
				AND
					dt < '$week.to'
			})
			$result.total($result.total+$result.$w.total)
		}
	}{
		^throw[bad.param;getStatHash;Wrong param]
	}

Re: как минимум от второго запроса можно точно избавиться

@getStatHash[week][w]
	^if($week){
		$result[^hash::create[]]
		^week.menu{
			$w($week.n)
			$result.$w[^oSql.hash{
				SELECT
					app_id
					,COUNT(*) AS cnt
				FROM
					weekly
				WHERE
					dt > '$week.from'
				AND
					dt < '$week.to'
				GROUP BY
					app_id
			}]
			$result.$w.total(0)
			^result.$w.foreach[k;v]{^result.$w.total.inc($v.cnt)}
			^result.total.inc($result.$w.total)
		}
	}{
		^throw[bad.param;getStatHash;Wrong param]
	}

какие вводные? структуру базы?

@getWeekTable[][tDays,week,day,from,to,n]
	$result[^table::create{n	from	to}]
	$tDays[^oSql.table{
#mysql specific request?
		SELECT
			DATE_FORMAT(dt,'%Y-%m-%d') as date
		FROM
			weekly
		GROUP BY
			date
		ORDER BY
			date
	}]
	^if($tDays){
		^tDays.menu{
			$day[^date::create[$tDays.date]]
			$week[^date:calendar[rus]($day.year;$day.month;$day.day)]
			$to[${week.year}-${week.month}-${week.day}]
			^if(!^from.length[]){ $from[$to] }
			^if($to gt $from){
				$n($n+1)
				^result.append[$n	$from	$to]
				$from[$to]
			}
		}
		^week.offset(-1)
		^result.append[^eval($n+1)	$from	${week.year}-${week.month}-${week.day} 23:59:59]
		^result.sort($result.n)[desc]
	}

-- phpMyAdmin SQL Dump
-- version 2.11.11.3
-- http://www.phpmyadmin.net
--
-- Хост: localhost
-- Время создания: Мар 14 2011 г., 07:15
-- Версия сервера: 4.1.25
-- Версия PHP: 4.4.9

SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";

--
-- База данных: `m9a`
--

-- --------------------------------------------------------

--
-- Структура таблицы `weekly`
--

CREATE TABLE IF NOT EXISTS `weekly` (
  `id` int(8) NOT NULL auto_increment,
  `dt` datetime NOT NULL default '0000-00-00 00:00:00' COMMENT 'when launch detected',
  `computer` varchar(127) NOT NULL default '' COMMENT 'where launched',
  `app_id` int(4) NOT NULL default '0' COMMENT 'application id from app table',
  PRIMARY KEY  (`id`),
  KEY `idx2` (`app_id`,`dt`),
  KEY `dt` (`dt`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 COMMENT='data for weekly report' AUTO_INCREMENT=38308 ;

Re: уже лучше

	$result.$w[^oSql.hash{
	SELECT
		app_id
		,COUNT(*) AS cnt
	FROM
		weekly
	WHERE
		WEEK(dt) IN (^week.menu{$week.number}[,])
	GROUP BY
		app_id
}]

Номеров недель всего 54.

собственно, если нужен образец самих данных, то вот они http://m9a.nafronte.ru/week.html

словами