SQL 查找日期列中间隔的查询

Question

我有一个 table，其中包含每天的状态和日期，我试图找出状态何时发生变化，以及每次状态变化之间是否存在差距/某种状态有多少天。

+------+----------------+--------+
| user | date           | status |
+----- +----------------+--------+
|    1 | 12/01/2017     | open   |
|    1 | 12/02/2017     | open   |
|    1 | 12/03/2017     | open   |
|    1 | 12/04/2017     | closed |
|    1 | 12/05/2017     | closed |
|    1 | 12/06/2017     | open   |
|    1 | 12/07/2017     | open   |
+------+----------------+--------+

预期输出：

+------+------------+----------------+-------- ----+------------+
| user | status     | days_in_status | min         | max        |
+----- +------------+----------------+-------------+------------+
|    1 | open       | 3              | 12/01/2017  | 12/03/2017 |
|    1 | closed     | 2              | 12/04/2017  | 12/05/2017 |
|    1 | open       | 2              | 12/06/2017  | 12/07/2017 |
+------+------ -----+----------------+-------------+-- ---------+

Answer 1

这是一种缺口孤岛问题。在这种情况下，从每一天中减去一个序号可能是识别“孤岛”的最简单解决方案：

select user, status, count(*) as num_days, min(date), max(date)
from (select t.*,
             row_number() over (partition by user, status order by date) as seqnum
      from t
     ) t
group by user, status, date - seqnum * interval '1 day'

Answer 2

在 Postgres 中，您可以直接从 date 中减去 integer：

SELECT "user", status
     , count(*) AS days_in_status, min(date) AS min, max(date) AS max
FROM  (
   SELECT "user", status, date
        , date - row_number() OVER (PARTITION BY "user", status ORDER BY date)::int AS grp_date
   FROM   tbl
   ) t
GROUP BY "user", status, grp_date
ORDER BY "user", min;

db<>fiddle here

参见：

Select longest continuous sequence
How do I determine the last day of the previous month using PostgreSQL?

旁白：user 是一个 reserved word，您应该避免将其作为实际列名。

Answer 3

SQL 查询以查找日期列中日期 GAPS 的宽度

从 Gordon 上面的答案中借用的完整示例显示了岛屿宽度和岛屿间隙，复制并粘贴到 postgresql 和运行中。

drop table if exists foobar; 
CREATE TABLE foobar( tick text, date_val date ); 

insert into foobar values('XYZ', '2021-01-03');  --island 1 has width 2
insert into foobar values('XYZ', '2021-01-04');  --island 1
insert into foobar values('XYZ', '2021-05-09');  --island 2 has width 3
insert into foobar values('XYZ', '2021-05-10');  --island 2 
insert into foobar values('XYZ', '2021-05-11');  --island 2
insert into foobar values('XYZ', '2021-07-07');  --island 3 has width 4
insert into foobar values('XYZ', '2021-07-08');  --island 3
insert into foobar values('XYZ', '2021-07-09');  --island 3
insert into foobar values('XYZ', '2021-07-10');  --island 3 
insert into foobar values('XYZ', '2022-10-10');  --island 4 has width 1

select tick, island_width, min_val, max_val, 
       min_val - lag(max_val) over (order by max_val) as gap_width from  
( 
  select tick, count(*) as island_width, 
         min(date_val) min_val, max(date_val) max_val 
  from ( 
    select t.*, 
    row_number() over ( partition by tick order by date_val ) as seqnum 
    from foobar t where tick = 'XYZ' 
    ) t 
  group by tick, date_val - seqnum * interval '1 day' 
) t2 order by max_val desc

打印：

┌──────┬──────────────┬────────────┬────────────┬───────────┐ 
│ tick │ island_width │  min_val   │  max_val   │ gap_width │ 
├──────┼──────────────┼────────────┼────────────┼───────────┤ 
│ XYZ  │            1 │ 2022-10-10 │ 2022-10-10 │       457 │ 
│ XYZ  │            4 │ 2021-07-07 │ 2021-07-10 │        57 │ 
│ XYZ  │            3 │ 2021-05-09 │ 2021-05-11 │       125 │ 
│ XYZ  │            2 │ 2021-01-03 │ 2021-01-04 │         ¤ │ 
└──────┴──────────────┴────────────┴────────────┴───────────┘

详情：在这个数据中有4个岛屿。 4个岛之间有3个缺口。最后一个岛有缺口。第一个岛有一个零间隙。

SQL 通过减去一天并递减 rownumber 直到所有日期都被计算在内，找到每个 min_date 和 max_date 之间的天数。 partition 以某种方式按日期和 2 天移动 window 使用 over 和 lag 以及默认后视 1. Black Box Demon magic 将岛屿分组测试通过。

在演练的帮助下：https://bertwagner.com/posts/gaps-and-islands

SQL 查找日期列中间隔的查询

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SQL 查询以查找日期列中日期 GAPS 的宽度