计算 False 发生之前每天 True 标志的连续数量

Question

您能帮我完成以下 SQL 任务吗？这是一个数据表：

|     date      |   hour   |   flag   |
|---------------|----------|----------|
|   2024-04-13  |    23    |   True   |
|   2024-04-13  |    22    |   True   |
|   2024-04-13  |    21    |   True   |
|   2024-04-13  |    20    |   False  |
|   2024-04-13  |    19    |   True   |
|   2024-04-13  |    18    |   True   |
|-------------------------------------|
|   2024-04-12  |    20    |   False  |
|   2024-04-12  |    19    |   True   |
|   2024-04-12  |    18    |   True   |
|   2024-04-12  |    17    |   True   |
|   2024-04-12  |    16    |   True   |
|   2024-04-12  |    15    |   True   |
|-------------------------------------|
|   2024-04-11  |    22    |   True   |
|   2024-04-11  |    18    |   True   |
|   2024-04-11  |    15    |   False  |
|   2024-04-11  |    10    |   True   |
|   2024-04-11  |     9    |   False  |
|   2024-04-11  |     8    |   True   |
|-------------------------------------|
|   2024-04-10  |    10    |   True   |
|   2024-04-10  |    9     |   True   |
|   2024-04-10  |    6     |   True   |
|   2024-04-10  |    3     |   False  |
|-------------------------------------|

我需要计算的是，每天，在出现任何假标志之前（从顶部开始）有多少个连续的真标志。所以，结果应该是这样的：

|     date      |   count  |
|---------------|----------|
|   2024-04-13  |     3    |
|   2024-04-12  |     0    |
|   2024-04-11  |     2    |
|   2024-04-10  |     3    |

注释：小时按降序排列，但可能存在漏洞，例如2024-04-11（结果仍然是 2，我们正在计算行数）。

提前谢谢您。

Answer 1

也许可以改进，但这确实有效。

SET @count := 0, @aux := 0, @prevdate := null, @end := 0, @prevend := 0;

SELECT date, max(counter_final) 
FROM (
   SELECT 
      *,
      (@aux := IF(@prevdate = date, 1 , 0)) AS aux,
      (@count := IF(flag = true, IF(@aux = 0, 1, @count + 1), 0)) AS counter,
      (@prevdate := date) AS prevdate,
      (@prevend := IF(@aux = 0, 0, IF(@prevend = 1, 1,@end) )) AS prevend,
      (@end := IF(@count = 0 AND flag = 0,1,0)) AS end,
      IF(@end + @prevend = 0, @count, 0) AS counter_final
  FROM mytable 
  ORDER BY date DESC, hour DESC
) aux_table
GROUP BY date ORDER BY date DESC;

结果

date        count
2024-04-13  3
2024-04-12  0
2024-04-11  2
2024-04-10  3

完整示例：SQL Fiffdle

Answer 2

你也许可以做这样的事情：

SELECT  date, max(countOver) AS TruthFlags
FROM    (
    SELECT  COUNT(CASE WHEN flag = 'True' THEN 1 END) OVER(PARTITION BY Date ORDER BY hour DESC ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)
        * CASE WHEN COUNT(CASE WHEN flag = 'False' THEN 1 END) OVER(PARTITION BY Date ORDER BY hour DESC ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) > 0 THEN 0 ELSE 1 END AS countOver
    ,   *
    FROM
    (
        VALUES  (N'2024-04-13', 23, N'True')
        ,   (N'2024-04-13', 22, N'True')
        ,   (N'2024-04-13', 21, N'True')
        ,   (N'2024-04-13', 20, N'False')
        ,   (N'2024-04-13', 19, N'True')
        ,   (N'2024-04-13', 18, N'True')
        ,   (N'2024-04-12', 20, N'False')
        ,   (N'2024-04-12', 19, N'True')
        ,   (N'2024-04-12', 18, N'True')
        ,   (N'2024-04-12', 17, N'True')
        ,   (N'2024-04-12', 16, N'True')
        ,   (N'2024-04-12', 15, N'True')
        ,   (N'2024-04-11', 22, N'True')
        ,   (N'2024-04-11', 18, N'True')
        ,   (N'2024-04-11', 15, N'False')
        ,   (N'2024-04-11', 10, N'True')
        ,   (N'2024-04-11', 9, N'False')
        ,   (N'2024-04-11', 8, N'True')
        ,   (N'2024-04-10', 10, N'True')
        ,   (N'2024-04-10', 9, N'True')
        ,   (N'2024-04-10', 6, N'True')
        ,   (N'2024-04-10', 3, N'False')
    ) t (date,hour,flag)
) x
group by date

我的想法是创建一个“零”运行的错误标志总数，从最后一个小时到第一个小时。如果存在错误标志，则此表达式：

CASE WHEN COUNT(CASE WHEN flag = 'False' THEN 1 END) OVER(PARTITION BY Date ORDER BY hour DESC) > 0 THEN 0 ELSE 1 END

创建 0 值，否则创建 1。

然后只需将归零器乘以当前 True 标志的数量即可。

最后，我使用

max(countOver)

来查找聚合真实标记数的峰值。