Postgresql 左连接

Question

我有两张桌子cars和usage 。 我每月为某些汽车创建一次使用记录。 现在我想获得我保存的最新使用情况的不同汽车列表。

首先请看一下表格

cars:

| id | model       | reseller_id |
|----|-------------|-------------|
| 1  | Samand Sall | 324228      |
| 2  | Saba 141    | 92933       |

usages:

| id | car_id | year | month | gas |
|----|--------|------|-------|-----|
| 1  | 2      | 2020 | 2     | 68  |
| 2  | 2      | 2020 | 3     | 94  |
| 3  | 2      | 2020 | 4     | 33  |
| 4  | 2      | 2020 | 5     | 12  |

问题就在这里

• 我只需要年月的最新用法

我尝试了很多方法，但没有一个足够好。 因为有时这个查询让我得到一个非最新的使用记录。

SELECT * FROM cars AS c
LEFT JOIN
     (select *
      from usages
     ) u on (c.id = u.car_id)
order by u.gas desc

Answer 1

您可以在派生表中使用 DISTINCT ON 执行此操作：

SELECT * 
FROM cars AS c
  LEFT JOIN (
    select distinct on (u.car_id) *
    from usages u
    order by u.car_id, u.year desc, u.month desc
  ) lu on c.id = lu.car_id
order by u.gas desc;

Answer 2

我认为您需要 window function row_number 。 这是演示。

select
  id,
  model,
  reseller_id
from
(
  select
    c.id,
    model,
    reseller_id,
    row_number() over (partition by u.car_id order by u.id desc) as rn
  from cars c
  left join usages u
  on c.id = u.car_id
) subq
where rn = 1