SQL: 按方向过滤行

Question

我有一个 table，包含 2 列日期（时间戳）、状态（布尔值）。我有很多价值，例如：

| date                      | status    |
|-------------------------- |--------   |
| 2018-11-05T19:04:21.125Z  | true      |
| 2018-11-05T19:04:22.125Z  | true      |
| 2018-11-05T19:04:23.125Z  | true      |
....

我需要得到这样的结果：

| date_from                 | date_to                   | status    |
|-------------------------- |-------------------------- |--------   |
| 2018-11-05T19:04:21.125Z  | 2018-11-05T19:04:27.125Z  | true      |
| 2018-11-05T19:04:27.125Z  | 2018-11-05T19:04:47.125Z  | false     |
| 2018-11-05T19:04:47.125Z  | 2018-11-05T19:04:57.125Z  | true      |

因此，我需要过滤所有 "same" 值并仅进入 return 状态 true/false。

我这样创建查询：

SELECT max("current_date"), current_status, previous_status
FROM (SELECT date as "current_date",
             status as current_status,
             (lag(status, 1) OVER (ORDER BY msgtime))::boolean AS previous_status
      FROM "table" as table
      ) as raw_data
group by current_status, previous_status

但作为回应，我只得到不超过 4 个值

Answer 1

是的，您可以使用 LAG，但您还需要一个运行计数器，每次状态更改时都会递增：

WITH cte1 AS (
    SELECT date, status, CASE WHEN LAG(status) OVER (ORDER BY date) = status THEN 0 ELSE 1 END AS chg
    FROM yourdata
), cte2 AS (
    SELECT date, status, SUM(chg) OVER (ORDER BY date) AS grp
    FROM cte1
)
SELECT MIN(date) AS date_from, MAX(date) AS date_to, status
FROM cte2
GROUP BY grp, status
ORDER BY date_from

DB Fiddle

Answer 2

这是一个缺口和孤岛问题。一个典型的方法是使用行号的差异：

select min(date), max(date), status
from (select t.*,
             row_number() over (order by date) as seqnum,
             row_number() over (partition by status order by date) as seqnum_s
      from t
     ) t
group by status, (seqnum - seqnum_s);

SQL: 按方向过滤行

SQL: FIlter rows by direction

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postgresql

where

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