获取进程在 Jinja 上每个阶段的时间间隔 SQL

Question

我需要获取进程在每个阶段的时间间隔，并考虑进程何时返回到某个阶段。例如：

stage_name	from_day	to_day
A	1	2
B	2	3
B	3	4
B	4	5
C	5	6
B	6	7
D	7

进程目前在阶段 D

我想要一个table这样的

stage_name	from_day	to_day
A	1	2
B	2	5
C	5	6
B	6	7
D	7

Answer 1

这是一个gaps and islands问题，您可以使用以下方法解决这个问题。

方法一

SELECT
    stage_name,
    MIN(from_day) as from_day,
    MAX(to_day) as to_day
FROM (
    SELECT
        *,
       ROW_NUMBER() OVER (ORDER BY from_day) - ROW_NUMBER() OVER (PARTITION  BY stage_name ORDER BY from_day) as grp
        FROM
            my_table
    ) t1

GROUP BY stage_name,grp
ORDER BY from_day, stage_name;

stage_name	from_day	to_day
A	1	2
B	2	5
C	5	6
B	6	7
D	7

View working demo on DB Fiddle

方法二

SELECT
    stage_name,
    MIN(from_day) as from_day,
    MAX(to_day) as to_day
FROM (
    SELECT
        *,
       SUM(same) OVER (PARTITION BY stage_name ORDER BY from_day) as grp
    FROM (
        SELECT
            *,
            CASE 
                WHEN from_day = LAG(to_day,1,from_day) OVER (PARTITION BY stage_name ORDER BY from_day) THEN 0
                ELSE 1
            END as same
        FROM
            my_table
    ) t1
) t2
GROUP BY stage_name,grp
ORDER BY from_day, stage_name;

stage_name	from_day	to_day
A	1	2
B	2	5
C	5	6
B	6	7
D	7

View working demo on DB Fiddle

上面的示例使用 from_day = LAG(to_day,1,from_day) OVER (PARTITION BY stage_name ORDER BY from_day) 来确定当前行 from_day 是否与相同 stage_name group/partition 中的前几行 to_day 相同from_day 订购。如果相同，则分配 0，否则分配 1。此子查询的输出已包含在下面供您阅读：

SELECT
    *,
    CASE 
        WHEN from_day = LAG(to_day,1,from_day) OVER (PARTITION BY stage_name ORDER BY from_day) THEN 0
        ELSE 1
    END as same
FROM
    my_table
ORDER BY from_day, stage_name;

stage_name	from_day	to_day	same
A	1	2	0
B	2	3	0
B	3	4	0
B	4	5	0
C	5	6	0
B	6	7	1
D	7		0

然后使用window函数SUM(same) OVER (PARTITION BY stage_name ORDER BY from_day)求出这些差异的累加和，从而创建组：

    SELECT
        *,
        SUM(same) OVER (PARTITION BY stage_name ORDER BY from_day) as grp
    FROM (
            SELECT
                *,
                CASE 
                    WHEN from_day = LAG(to_day,1,from_day) OVER (PARTITION BY stage_name ORDER BY from_day) THEN 0
                    ELSE 1
                END as same
            FROM
                my_table
        ) t1
     ORDER BY from_day, stage_name;

stage_name	from_day	to_day	same	grp
A	1	2	0	0
B	2	3	0	0
B	3	4	0	0
B	4	5	0	0
C	5	6	0	0
B	6	7	1	1
D	7		0	0

最后，通过按 stage_name 和 grp 分组，我们可以在每个组中找到所需的值，在本例中最早的 from_day 使用 MIN 和最新的to_day 使用 MAX

获取进程在 Jinja 上每个阶段的时间间隔 SQL

Get time intervals where the process was on each stage on Jinja SQL

mysql

sql

intervals

gaps-and-islands

方法一

方法二