Question

我有三张桌子：

user(id, name);

tasks(id, user_id, text, date);

bonus(id, user_id, sum, date, type).

我正在尝试编写一个查询，该列表将为我提供所有用户的列表，他们的任务数量以及每种类型的奖金总额，每个都在单独的列中。

有三种类型的奖金：

type = (1, 2, 3)

所以，我的意思是一张表，看起来像这样

name | task_count | bonus_1 | bonus_2 | bonus_3

Answer 1

SELECT u.name, 
        t.task_count, 
        SUM(CASE WHEN b.type=1 then b.sum END) AS bonus_1,
        SUM(CASE WHEN b.type=2 then b.sum END) AS bonus_2,
        SUM(CASE WHEN b.type=3 then b.sum END) AS bonus_3
FROM user u 
LEFT JOIN (SELECT user_id, COUNT(*) AS task_count 
           FROM tasks 
           GROUP BY user_id) t ON t.user_id = u.id
LEFT JOIN bonus b ON b.user_id = u.id 
GROUP BY u.id,u.name,t.task_count

Answer 2

注意：以下操作无效！保留信息

SELECT
    user.name,
    COUNT(tasks.id) AS task_count,
    SUM(bonus1.sum) AS bonus_1,
    SUM(bonus2.sum) AS bonus_2,
    SUM(bonus3.sum) AS bonus_3
FROM user
    LEFT OUTER JOIN tasks ON tasks.user_id = user.id
    LEFT OUTER JOIN bonus1 ON bonus1.user_id = user.id
        AND bonus1.type = 1
    LEFT OUTER JOIN bonus2 ON bonus2.user_id = user.id
        AND bonus1.type = 2
    LEFT OUTER JOIN bonus3 ON bonus3.user_id = user.id
        AND bonus1.type = 3
GROUP BY
    user.name
ORDER BY
    user.name

如果不是每个用户总是至少有一个任务和至少一种类型的奖励，则可能需要特别处理NULL值。

Answer 3

在加入user之前，请考虑进行聚合。例如：

select  u.name
,       t.task_count
,       b.bonus_1
,       b.bonus_2
,       b.bonus_3
,       t.comment_count
from    user u
left join
        (
        select  user_id
        ,       count(distinct t.id) as task_count
        ,       count(distinct c.id) as comment_count
        from    tasks t
        left join
                comment c
        on      c.task_id = t.id
        group by
                t.user_id
        ) t
on      t.user_id = u.id
left join
        (
        select  user_id
        ,       sum(case when type = 1 then sum end) as bonus_1
        ,       sum(case when type = 2 then sum end) as bonus_3
        ,       sum(case when type = 3 then sum end) as bonus_2
        from    bonus
        group by
                user_id
        ) b
on      b.user_id = u.id

SQL查询（不带子查询）

3 个答案: