如何有SQL查询2级的子查询分为

Question

我具备这些表的数据库：

1. 用户（ID，电子邮件）
2. 旅行（ID，driver_id）
3. MatchedTrips（ID，trip_id ）

我需要为每个用户获得他创建的旅行总数除以找到的总匹配数。

我一直在为此构建原始SQL查询。这是我试过的，并且确定它远没有正确。

SELECT 
    users.email, 
    total_trips.count1/total_matches.count2 
FROM users CROSS JOIN (SELECT 
     users.email, 
     count(trips.driver_id) AS count1 
     FROM trips 
     INNER JOIN users ON trips.driver_id = users.id 
     GROUP BY users.email) total_trips 
     CROSS JOIN (SELECT users.email, count(matches.trip_id) AS count2 
        FROM matches 
        LEFT JOIN trips ON matches.trip_id = trips.id 
        LEFT JOIN users ON trips.driver_id = users.id 
        GROUP BY users.email) total_matches; 

Answer 1

最简单的方法可能是使用count(distinct)：

select u.email, 
     count(distinct t.id) as num_trips, 
     count(distinct m.id) as num_matches, 
     (count(distinct t.id)/count(distinct m.id)) as ratio 
from users u left join 
    trips t 
    on t.driver_id = u.id left join 
    matches m 
    on m.trip_id = t.trip_id 
group by u.email; 

注意：如果邮件是唯一的，那么查询可以简化。在某些情况下，count(distinct)可能很昂贵。

Answer 2

你可以算出这样的方式为每个驱动程序的总人次及总比赛：

select driver_id, count(t.id) as total_trips, count(m.id) as total_matches 
from trips t 
left join matches m on (t.id = trip_id) 
group by 1 

使用该查询作为派生表与users加入：

select email, total_trips, total_matches, total_trips::dec/ nullif(total_matches, 0) result 
from users u 
left join (
    select driver_id, count(t.id) as total_trips, count(m.id) as total_matches 
    from trips t 
    left join matches m on (t.id = trip_id) 
    group by 1 
    ) s on u.id = driver_id 
order by 1; 

SQLFiddle.