带有 where 子句的 SQL Lag()

Posted 2023-03-31

【中文标题】带有 where 子句的 SQL Lag()

【英文标题】：SQL Lag() with where clause

【发布时间】：2021-02-24 18:34:11

【问题描述】：

我有一张这样的桌子-

row_no	Movie.	movie_start_time	movie_end_time
1	A	2021-02-01 01:00:00	2021-02-01 02:00:00
2	B	2021-02-01 01:00:00	2021-02-01 02:00:00
3	A	2021-02-01 01:30:00	2021-02-01 02:30:00
4	A	2021-02-01 01:30:00	2021-02-01 02:30:00
5	A	2021-02-01 02:15:00	2021-02-01 03:15:00
6	B	2021-02-01 02:15:00	2021-02-01 03:15:00
7	A	2021-02-01 04:15:00	2021-02-01 05:15:00

我想在表中添加一个额外的列，该列在同一部电影的先前播放时间和当前播放时间之间存在差异。我还有一个条件是之前的播放时间和当前的播放时间不应该重叠。所以在上面的场景中，结果应该如下所示：

row_no	Movie.	movie_start_time	movie_end_time.	last_play
1	A	2021-02-01 01:00:00	2021-02-01 02:00:00	-
2	B	2021-02-01 01:00:00	2021-02-01 02:00:00	-
3	A	2021-02-01 01:30:00	2021-02-01 02:30:00	-
4	A	2021-02-01 01:30:00	2021-02-01 02:30:00	-
5	A	2021-02-01 02:15:00	2021-02-01 03:15:00	15 minutes
6	B	2021-02-01 02:15:00	2021-02-01 03:15:00	15 minutes
7	A	2021-02-01 04:15:00	2021-02-01 05:15:00	60 minutes

我尝试编写以下查询以获取上一个电影结束时间，以便稍后计算差异：

select movie, movie_start_time, movie_end_time, lag(movie_end_time) over (partition by movie order by movie_start_time) prev_end_time from table where prev_end_time <= movie_start_time

但这不起作用，因为我们不能在窗口函数中添加 where 子句。有没有其他方法可以解决这个问题？

【问题讨论】：

select * from (<that query of yours>) t where prev_end_time <= movie_start_time?

你确定第五行是正确的吗？因为它与第四行的另一部电影 A 重叠。

@GSerg 这行不通，因为滞后只会给我上一行，我需要满足条件 prev_end_time 的上一行

@ARAT 我认为我选择的示例有点令人困惑 - 第 5 行 15 分钟是与不重叠的先前记录的区别 - 这是第 1 行

【参考方案1】：

所以我用postgreSQL建表：

CREATE TABLE IF NOT EXISTS table1 (
  row_no INT,
  Movie CHAR,
  movie_start_time timestamp,
  movie_end_time timestamp
);

并将记录插入到该表中：

INSERT INTO table1 (row_no, Movie, movie_start_time, movie_end_time) VALUES(1, 'A', '2021-02-01 01:00:00', '2021-02-01 02:00:00');
INSERT INTO table1 (row_no, Movie, movie_start_time, movie_end_time) VALUES(2, 'B', '2021-02-01 01:00:00',  '2021-02-01 02:00:00');
INSERT INTO table1 (row_no, Movie, movie_start_time, movie_end_time) VALUES(3, 'A', '2021-02-01 01:30:00',  '2021-02-01 02:30:00');
INSERT INTO table1 (row_no, Movie, movie_start_time, movie_end_time) VALUES(4, 'A', '2021-02-01 01:30:00', '2021-02-01 02:30:00');
INSERT INTO table1 (row_no, Movie, movie_start_time, movie_end_time) VALUES(5, 'A', '2021-02-01 02:15:00', '2021-02-01 03:15:00');
INSERT INTO table1 (row_no, Movie, movie_start_time, movie_end_time) VALUES(6, 'B', '2021-02-01 02:15:00', '2021-02-01 03:15:00');
INSERT INTO table1 (row_no, Movie, movie_start_time, movie_end_time) VALUES(7, 'A', '2021-02-01 04:15:00', '2021-02-01 05:15:00');

那么你想要的是：

select row_no, movie, movie_start_time, movie_end_time, EXTRACT(EPOCH FROM (movie_start_time - prev_end_time)::INTERVAL)/60 AS last_play FROM 
(select row_no,movie, movie_start_time, movie_end_time, lag(movie_end_time) over (partition by movie order by movie_start_time) AS "prev_end_time"
from table1) t
where prev_end_time <= movie_start_time
ORDER BY movie_start_time

【讨论】：

在您的结果中，预期结果中缺少第 5 行。您只是在检查该电影的上一个记录。但是我需要电影的前一个记录，其中 prev_end_time

【参考方案2】：

我能够通过以下查询解决问题：

with a as 
(select row_no,movie, movie_start_time, movie_end_time, 
array_agg(movie_end_time) over (partition by movie order by movie_start_time
 rows between unbounded preceding and 1 preceding) AS prev_end_time from `table1`),

b as (select row_no,movie, movie_start_time, movie_end_time,
case 
when prev_end_time is null then null
else (select max(i_prev_end_time) from unnest(prev_end_time)i_prev_end_time 
where i_prev_end_time <= movie_start_time)
end previous_end_time from a)

select row_no,movie, movie_start_time, movie_end_time,
 unix_seconds(movie_start_time) - unix_seconds(previous_end_time) last_run from b