获取最新的项目，即使从中为null

Question

我有三个不同的表，我尝试根据下面的不同表值获取输出，但是当我只想要最新的表时，它现在返回表等级的所有字段。 输出应如下所示：

desire output: ╔════╦════════╦═══════╦════════════╦═════════╗ ║ id ║ value ║ name ║ sum(total) ║ version ║ ╠════╬════════╬═══════╬════════════╬═════════╣ ║ 1 ║ hola ║ peter ║ 52 ║ second ║ ║ 2 ║ hello ║ john ║ 16 ║ third ║ ║ 3 ║ privet ║ luke ║ 3 ║ first ║ ╚════╩════════╩═══════╩════════════╩═════════╝

table users ╔════╦════════╦═══════╦══╗ ║ id ║ value ║ name ║ ║ ╠════╬════════╬═══════╬══╣ ║ 1 ║ hola ║ peter ║ ║ ║ 2 ║ hello ║ john ║ ║ ║ 3 ║ privet ║ luke ║ ║ ║ 4 ║ yay ║ eli ║ ║ ╚════╩════════╩═══════╩══╝

table props ╔════╦════════╦═════════╦══╗ ║ id ║ value ║ amount ║ ║ ╠════╬════════╬═════════╬══╣ ║ 1 ║ hola ║ 40 ║ ║ ║ 2 ║ hello ║ 5 ║ ║ ║ 3 ║ privet ║ 3 ║ ║ ║ 4 ║ hello ║ 2 ║ ║ ║ 5 ║ hello ║ 9 ║ ║ ║ 6 ║ hola ║ 12 ║ ║ ╚════╩════════╩═════════╩══╝ table ranks ╔════╦════════╦═════════╦════════════╗ ║ id ║ value ║ version ║ date ║ ╠════╬════════╬═════════╬════════════╣ ║ 1 ║ hola ║ first ║ 06-20-2015 ║ ║ 2 ║ hello ║ first ║ 04-24-2015 ║ ║ 3 ║ privet ║ first ║ 02-01-2015 ║ ║ 4 ║ hello ║ second ║ 10-05-2015 ║ ║ 5 ║ hello ║ third ║ 01-03-2016 ║ ║ 6 ║ hola ║ second ║ 01-23-2016 ║ ╚════╩════════╩═════════╩════════════╝ table ranks ╔════╦════════╦═════════╦════════════╗ ║ id ║ value ║ version ║ date ║ ╠════╬════════╬═════════╬════════════╣ ║ 1 ║ hola ║ first ║ 06-20-2015 ║ ║ 2 ║ hello ║ first ║ 04-24-2015 ║ ║ 3 ║ privet ║ first ║ 02-01-2015 ║ ║ 4 ║ hello ║ second ║ 10-05-2015 ║ ║ 5 ║ hello ║ third ║ 01-03-2016 ║ ║ 6 ║ hola ║ second ║ 01-23-2016 ║ ╚════╩════════╩═════════╩════════════╝

SELECT
  users.id,
  users.value,
  users.name,
  sum(props.amount),
  ranks.version

FROM
  users
join props
on (users.value=props.value)
left join
on (users.value = ranks.value)


group by
  users.id,
  users.values,
  users.title,
  ranks.version

`

Answer 1

首先，如果只需要一行，则不group by版本group by 。 使用聚合时， keep是一种有用的方法：

SELECT u.id, u.value, u.name,
       sum(p.amount),
       max(r.version) keep (dense_rank first order by r.version) as version,
       max(r.value) keep (dense_rank first order by r.version) as value
FROM users u join
     props p
     on u.value = p.value left join
     ranks r
     on u.value = r.value
GROUP BY u.id, u.values, u.title;