Question

我的表：

ID   NUM   VAL
1    1     Hello
1    2     Goodbye
2    2     Hey
2    4     What's up?
3    5     See you

如果我想返回每个ID的最大数量，那就非常干净了：

SELECT MAX(NUM) FROM table GROUP BY (ID)

但是如果我想获取与每个ID的每个数字的最大值相关联的值呢？

为什么我不能这样做：

SELECT MAX(NUM) OVER (ORDER BY NUM) FROM table GROUP BY (ID)

为什么会出错？我想让这个选择按ID分组，而不是为每个窗口单独分区......

编辑：错误是“不是GROUP BY表达式”。

Answer 1

您可以使用MAX() KEEP(DENSE_RANK LAST...)功能：

with sample_data as (
  select 1 id, 1 num, 'Hello' val from dual union all
  select 1 id, 2 num, 'Goodbye' val from dual union all
  select 2 id, 2 num, 'Hey' val from dual union all
  select 2 id, 4 num, 'What''s up?' val from dual union all
  select 3 id, 5 num, 'See you' val from dual)
select id, max(num), max(val) keep (dense_rank last order by num)
from sample_data
group by id;

Answer 2

使用窗口函数时，不再需要使用GROUP BY，这就足够了：

select id, 
     max(num) over(partition by id) 
from x

实际上你可以在不使用窗口函数的情况下得到结果：

select * 
from x
where (id,num) in
  (
     select id, max(num) 
     from x 
     group by id
  )

输出：

ID  NUM VAL
1   2   Goodbye
2   4   What's up
3   5   SEE YOU

http://www.sqlfiddle.com/#!4/a9a07/7

如果您想使用窗口功能，可以这样做：

select id, val, 
     case when num =  max(num) over(partition by id) then
        1
     else
        0
     end as to_select
from x 
where to_select = 1

或者这个：

select id, val 
from x 
where num =  max(num) over(partition by id)

但由于不允许这样做，你必须这样做：

with list as
(
  select id, val, 
     case when num =  max(num) over(partition by id) then
        1
     else
        0
     end as to_select
  from x
)
select * 
from list 
where to_select = 1

http://www.sqlfiddle.com/#!4/a9a07/19

Answer 3

如果您希望获得包含MAX(num) GROUP BY id值的行，这往往是一种常见模式......

WITH
  sequenced_data
AS
(
  SELECT
    ROW_NUMBER() OVER (PARTITION BY id ORDER BY num DESC) AS sequence_id,
    *
  FROM
    yourTable
)
SELECT
  *
FROM
  sequenced_data
WHERE
  sequence_id = 1

修改

我不知道TeraData是否允许这样做，但逻辑似乎有意义......

SELECT * FROM yourTable WHERE num = MAX(num) OVER (PARTITION BY id)

或者也许......

SELECT * FROM ( SELECT *, MAX(num) OVER (PARTITION BY id) AS max_num_by_id FROM yourTable ) AS sub_query WHERE num = max_num_by_id

这与我之前的回答略有不同;如果多个记录与同一个MAX(num)绑定，则会返回所有记录，另一个答案将只返回一个。

修改

在您提议的SQL中，错误与OVER()子句包含不在GROUP BY中的字段这一事实有关。这就像试图这样做......

SELECT id, num FROM yourTable GROUP BY id

num无效，因为对于返回的每一行，该字段中可能有多个值（返回的行由GROUP BY id定义）。

以同样的方式，您不能将num放在OVER()子句中。

SELECT id, MAX(num), <-- Valid as it is an aggregate MAX(num) <-- still valid OVER(PARTITION BY id), <-- Also valid, as id is in the GROUP BY MAX(num) <-- still valid OVER(PARTITION BY num) <-- Not valid, as num is not in the GROUP BY FROM yourTable GROUP BY id

如果您无法在OVER()子句中指定内容，并且（我认为）何时可以显示答案，请参阅此问题：over-partition-by-question

Oracle SQL - 分析功能超过一个组？

3 个答案: