Question

我有一个包含这些表的数据库：

用户（身份证，电子邮件）
旅行（id，driver_id）
MatchedTrips（id，trip_id）

我需要为每个用户他创建的旅行总数除以找到的总匹配数。

我坚持为此构建原始SQL查询。这是我尝试过的，并且确定它远非正确。

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

您可以通过以下方式计算每位车手的总行程和总比赛：

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

将此查询用作与users联接的派生表：

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.

Answer 2

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：如果电子邮件是唯一的，则可以简化查询。 count(distinct)在某些情况下可能会很昂贵。

如何将SQL查询与2个子查询分开

2 个答案: