Question

是否可以结合窗口函数OVER(PARTITION BY id)计算不同的值？目前我的查询如下：

SELECT congestion.date, congestion.week_nb, congestion.id_congestion,
   congestion.id_element,
ROW_NUMBER() OVER(
    PARTITION BY congestion.id_element
    ORDER BY congestion.date),
COUNT(DISTINCT congestion.week_nb) OVER(
    PARTITION BY congestion.id_element
) AS week_count
FROM congestion
WHERE congestion.date >= '2014.01.01'
AND congestion.date <= '2014.12.31'
ORDER BY id_element, date

但是，当我尝试执行查询时，出现以下错误：

"COUNT(DISTINCT": "DISTINCT is not implemented for window functions"

Answer 1

不，正如错误消息所述，DISTINCT未使用Windows函数实现。将this link中的信息附加到您的案例中，您可以使用以下内容：

WITH uniques AS (
 SELECT congestion.id_element, COUNT(DISTINCT congestion.week_nb) AS unique_references
 FROM congestion
WHERE congestion.date >= '2014.01.01'
AND congestion.date <= '2014.12.31'
 GROUP BY congestion.id_element
)

SELECT congestion.date, congestion.week_nb, congestion.id_congestion,
   congestion.id_element,
ROW_NUMBER() OVER(
    PARTITION BY congestion.id_element
    ORDER BY congestion.date),
uniques.unique_references AS week_count
FROM congestion
JOIN uniques USING (id_element)
WHERE congestion.date >= '2014.01.01'
AND congestion.date <= '2014.12.31'
ORDER BY id_element, date

根据具体情况，您还可以将子查询直接放入SELECT - 列表：

SELECT congestion.date, congestion.week_nb, congestion.id_congestion,
   congestion.id_element,
ROW_NUMBER() OVER(
    PARTITION BY congestion.id_element
    ORDER BY congestion.date),
(SELECT COUNT(DISTINCT dist_con.week_nb)
    FROM congestion AS dist_con
    WHERE dist_con.date >= '2014.01.01'
    AND dist_con.date <= '2014.12.31'
    AND dist_con.id_element = congestion.id_element) AS week_count
FROM congestion
WHERE congestion.date >= '2014.01.01'
AND congestion.date <= '2014.12.31'
ORDER BY id_element, date

Answer 2

我发现最简单的方法是使用子查询/ CTE和条件聚合：

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Answer 3

使分区集更小，直到计数字段没有重复为止：

SELECT congestion.date, congestion.week_nb, congestion.id_congestion,
   congestion.id_element,
ROW_NUMBER() OVER(
    PARTITION BY congestion.id_element
    ORDER BY congestion.date),
COUNT(congestion.week_nb) -- remove distinct 
OVER(
    PARTITION BY congestion.id_element,
                 -- add new fields which will restart counter in case duplication
                 congestion.id_congestion
) AS week_count
FROM congestion
WHERE congestion.date >= '2014.01.01'
AND congestion.date <= '2014.12.31'
ORDER BY id_element, date

Answer 4

由于这是从Google弹出的第一个结果，因此我将添加此可重复的示例，类似于戈登的答案：

让我们首先创建一个示例表：

WITH test as 
(
SELECT * 
FROM (VALUES
(1, 'A'),
(1, 'A'),
(2, 'B'),
(2, 'B'),
(2, 'D'),
(3, 'C'),
(3, 'C'),
(3, 'C'),
(3, 'E'),
(3, 'F')) AS t (id_element, week_nb)
)

select * from test

这将产生：

id_element week_nb
1   A
1   A
2   B
2   B
2   D
3   C
3   C
3   C
3   E
3   F

然后，执行以下操作：

select 
  id_element,
  week_nb,
  sum(first_row_in_sequence) over (partition by id_element) as distinct_week_nb_count
from 
(
select 
  id_element,
  week_nb,
  case when row_number() over (partition by id_element, week_nb) = 1 then 1 else 0 end as first_row_in_sequence
from test
) as sub

收益

id_element week_nb distinct_week_nb_count
1   A   1
1   A   1
2   B   2
2   B   2
2   D   2
3   C   3
3   C   3
3   C   3
3   E   3
3   F   3

使用OVER（PARTITION BY id）计算不同的值

4 个答案: