группа mysql по одному столбцу, но агрегируется на нескольких столбцах?

Question

Ниже приведен большой запрос, он работает для 4 столбцов (4 виртуальных стола), но не для 12 столбцов, что было бы лучшим способом сделать это.

Ниже запрос для 4 колонок,

select tbl1.count1, tbl2.fresh1, tbl3.new1, tbl4.score1, tbl1.user_id from 
(
    (select count(id) as count1, user_id 
     from user_ranking 
     where rank = 1 
     group by user_id 
    ) as tbl1 
JOIN 
    (select count(id) as fresh1, user_id 
     from user_ranking 
     where rank = 1 and is_fresh = 1 
     group by user_id 
    ) as tbl2 ON tbl1.user_id = tbl2.user_id 
JOIN 
    (select count(id) as new1, user_id 
     from user_ranking 
     where rank = 1 and is_new = 1 
     group by user_id 
    ) as tbl3 ON tbl2.user_id = tbl3.user_id 
JOIN 
    (select AVG(score) as score1, user_id 
     from user_ranking 
     where rank = 1 group by user_id 
    ) as tbl4 ON tbl3.user_id = tbl4.user_id 
); 

Над отлично работает для меня, но не работает на 12 столбцов.

Как было предложено в ответ ниже я попытался

select user_id, SUM(rank = 1) as count1, SUM(rank = 1 and is_fresh = 1) as fresh1, SUM(rank = 1 and is_new = 1) as new1, 
SUM(IF(rank=1, score, 0))/SUM(rank=1) as score1 from user_ranking group by user_id; 

Это дает другой результат, что выше 4 столбца запроса.

Ниже запрос для 12 столбца, который не работает

select tbl1.user_id, 
tbl1.count1, tbl2.fresh1, tbl3.new1, tbl4.score1, 
tbl5.count2, tbl6.fresh2, tbl7.new2, tbl8.score2, 
tbl9.count3, tbl10.fresh3, tbl11.new3, tbl12.score3 
from 
(
    (select count(id) as count1, user_id 
     from user_ranking 
     where rank = 1 
     group by user_id 
    ) as tbl1 
JOIN 
    (select count(id) as fresh1, user_id 
     from user_ranking 
     where rank = 1 and is_fresh = 1 
     group by user_id 
    ) as tbl2 
JOIN 
    (select count(id) as new1, user_id 
     from user_ranking 
     where rank = 1 and is_new = 1 
     group by user_id 
    ) as tbl3 
JOIN 
    (select AVG(score) as score1, user_id 
     from user_ranking 
     where rank = 1 group by user_id 
    ) as tbl4 
JOIN 
    (select count(id) as count2, user_id 
     from user_ranking 
     where rank = 2 
     group by user_id 
    ) as tbl5 
JOIN 
    (select count(id) as fresh2, user_id 
     from user_ranking 
     where rank = 2 and is_fresh = 1 
     group by user_id 
    ) as tbl6 
JOIN 
    (select count(id) as new2, user_id 
     from user_ranking 
     where rank = 2 and is_new = 1 
     group by user_id 
    ) as tbl7 
JOIN 
    (select AVG(score) as score2, user_id 
     from user_ranking 
     where rank = 2 group by user_id 
    ) as tbl8 
JOIN 
    (select count(id) as count3, user_id 
     from user_ranking 
     where rank = 3 
     group by user_id 
    ) as tbl9 
JOIN 
    (select count(id) as fresh3, user_id 
     from user_ranking 
     where rank = 3 and is_fresh = 1 
     group by user_id 
    ) as tbl10 
JOIN 
    (select count(id) as new3, user_id 
     from user_ranking 
     where rank = 4 and is_new = 1 
     group by user_id 
    ) as tbl11 
JOIN 
    (select AVG(score) as score3, user_id 
     from user_ranking 
     where rank = 3 group by user_id 
    ) as tbl12  
); 

Таблица схемы

CREATE TABLE IF NOT EXISTS `user_ranking` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `user_id` int(11) NOT NULL, 
    `rank` int(11) NOT NULL, 
    `is_fresh` int(11) NOT NULL, 
    `is_new` int(11) NOT NULL, 
    `score` int(11) NOT NULL, 
    PRIMARY KEY (`id`) 
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ; 

-- 
-- Dumping data for table `user_ranking` 
-- 

INSERT INTO `user_ranking` (`id`, `user_id`, `rank`, `is_fresh`, `is_new`, `score`) VALUES 
(1, 1, 1, 1, 0, 25), 
(2, 2, 1, 1, 0, 67), 
(3, 1, 2, 0, 1, 34), 
(4, 2, 2, 0, 1, 24), 
(5, 2, 1, 0, 1, 36), 
(6, 1, 2, 1, 1, 1), 
(7, 1, 1, 0, 1, 75); 

Answer 1

просто переместить состояние от каждого ГДЕ SUM

SELECT 
    user_id, 
    SUM(rank = 1) AS rank1, 
    SUM(rank = 2) AS rank2, 
    SUM(rank = 2 and is_fresh = 1) AS rank2_fresh, 
    SUM(IF(rank = 1, age, 0))/SUM(rank = 1) AS rank_1_avg_age 
    ... 
FROM users 
GROUP BY user_id