在 PostgreSQL 中使用 window 函数是否可以找出一行中的项目数？

Question

如何找出有多少卖家有付款，且连续付款时间小于1分钟，且连续至少执行3次？（答案是 2 个卖家）以及如何计算此类付款的次数？（答案是 10 次付款）貌似这样的问题可以用window函数解决，不过我没遇到过这种问题

CREATE TABLE T (seller_id int, payment_id varchar(3), payment_time timestamp, second_diff int);
    
INSERT INTO T (seller_id, payment_id, payment_time, second_diff)
VALUES
    (1, 'pl',  '2015-01-08 09:23:04', 151),
    (1, 'p2',  '2015-01-08 09:25:35', 50),
    (1, 'p3',  '2015-01-08 09:26:25', 48),
    (1, 'p4',  '2015-01-08 09:27:23', 36),
    (1, 'p5',  '2015-01-08 09:27:59', 41),
    (1, 'p6',  '2015-01-08 09:28:40', 70),
    (1, 'p7',  '2015-01-08 09:29:50', 50),
    (1, 'p8',  '2015-01-08 09:30:40', 45),
    (1, 'p9',  '2015-01-08 09:31:25', 35),
    (1, 'p10', '2015-01-08 09:32:00', null),
    (2, 'pll', '2015-01-08 09:25:35', 25),
    (2, 'p12', '2015-01-08 09:26:00', 55),
    (2, 'p13', '2015-01-08 09:26:55', 30),
    (2, 'p14', '2015-01-08 09:27:25', 95),
    (2, 'p15', '2015-01-08 09:29:00', null),
    (3, 'p16', '2015-01-08 10:41:00', 65),
    (3, 'p17', '2015-01-08 10:42:05', 75),
    (3, 'p18', '2015-01-08 10:43:20', 90),
    (3, 'p19', '2015-01-08 10:43:20', 39),
    (3, 'p20', '2015-01-08 10:43:59', 50),
    (3, 'p21', '2015-01-08 10:44:49', null);

Answer 1

使用包含整数除以 60 的 OVER 子句的聚合函数，然后对该除法进行过滤以获得结果 0

WITH T AS
(
SELECT ... COUNT(*) OVER(PARTITION BY second_diff / 60) AS CNT, second_diff / 60 AS GRP
FROM ....
) 
SELECT * FROM T
WHERE GRP = 0

Answer 2

with A as (
    select seller_id, payment_time, second_diff,
        case when
            lag(case when second_diff < 60 then 1 else 0 end)
                over (partition by seller_id order by payment_time)
              = case when second_diff < 60 then 1 else 0 end
            then 0 else 1 end as transition
    from T
), B as (
    select *,
        sum(transition)
            over (partition by seller_id order by payment_time) as grp
    from A
), C as (
    select seller_id, count(*) as p
    from B
    where second_diff < 60
    group by seller_id, grp
    having count(*) >= 3
) 
select count(distinct seller_id) as sellers, sum(p) as payments
from C;

此方法会查找值中的转换，并对其进行计数。内部 case 表达式的输出值并不重要，只要它们匹配即可。

https://dbfiddle.uk/?rdbms=postgres_9.6&fiddle=606b796d793248336a95637f02ce117b

以下是主题的一些变体：

选项 #1b:

with A as (
    select seller_id, payment_time, second_diff,
        case when
            lag(case when second_diff < 60 then 1 else 0 end)
                over (partition by seller_id order by payment_time)
              = case when second_diff < 60 then 1 else 0 end
            then 0 else 1 end as transition
    from T
), B as (
    select *,
        sum(transition)
            over (partition by seller_id order by payment_time) as grp
    from A
)
select
    dense_rank() over (order by seller_id)
      + dense_rank() over (order by seller_id desc) - 1 as sellers,
    sum(count(*)) over () as payments
from B
where second_diff < 60
group by seller_id, grp
having count(*) >= 3
limit 1;

这只是一步完成 count(distinct) 的另一种方法。

选项#2：

with A as (
    select seller_id, payment_time, second_diff,
        row_number() over (partition by seller_id order by payment_time) as rn
    from T
), B as (
    select *,
        rn - row_number() over (partition by seller_id order by payment_time) as grp
    from A
    where second_diff < 60    
), C as (
    select seller_id, count(*) as p
    from B
    group by seller_id, grp
    having count(*) >= 3
) 
select count(distinct seller_id) as sellers, sum(p) as payments
from C;

此方法使用行编号预过滤和 post 过滤在系列中查找中断。

在 PostgreSQL 中使用 window 函数是否可以找出一行中的项目数？

Is it possible to find out the number of items in a row by using window functions in PostgreSQL?

sql

postgresql

analytical

window-functions

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