间隙和孤岛 - 如何按 ID 对每组连续行求和

Question

下面是我当前的 SQL 代码和输出。我只需要获得 CD 等于 STG（以黄色突出显示）的连续（或单个）行的 EFF_DAYS 的总和。

    SELECT ROW_NUMBER() OVER (PARTITION BY ID ORDER BY TMSP, EFF_DT) RN, 
Z2.* 
FROM (
    SELECT CASE WHEN (LAG_CD IS NULL OR LAG_CD NOT IN ('STG')) AND CD IN ('STG') 
    THEN RANK() OVER (PARTITION BY ID ORDER BY TMSP, EFF_DT)

    WHEN CD = LAG_CD AND CD IN ('STG') 
    THEN RANK() OVER (PARTITION BY ID ORDER BY TMSP, EFF_DT)

    WHEN CD = LAG_CD AND CD != LEAD_CD 
    THEN RANK() OVER (PARTITION BY ID ORDER BY TMSP, EFF_DT)

    END AS CASES, 
    Z.* FROM (
                SELECT ID,
                LAG(CD) OVER (PARTITION BY ID ORDER BY TMSP, EFF_DT) AS LAG_CD,
                LEAD(CD) OVER (PARTITION BY ID ORDER BY TMSP, EFF_DT) AS LEAD_CD,
                CD, 
                TMSP, 
                EFF_DT, 
                END_EFF_DT,
                DATEDIFF(day, EFF_DT, END_EFF_DT) AS EFF_DAYS
        FROM #POSTCHG_ROWS 
        WHERE ID IN ('ABC123', 'XYZ789')
        ) Z
    ) Z2 ORDER BY TMSP, EFF_DT

我已经尝试了各种行号和排名，但我似乎无法使 CASES 列正确。我花了几个小时研究其他 gap-island sql 解决方案，但没有遇到下面的确切情况。

理想情况下，我的 CASES 列会像下面这样输出，这样我就可以按 CASES、ID、连续行块的起始 TMSP 进行分组，然后计算：SUM(EFF_DAYS)。

下面是我的目标输出：

Answer 1

您只对一系列相邻的“CTG”行感兴趣。我认为最简单的方法是 window 非“STG”值的计数来定义组，然后过滤和聚合：

select 
    id,
    min(tmsp) tmsp,
    min(eff_dt) eff_dt,
    sum(datediff(day, eff_dt, end_eff_dt)) sum_eff_days
from (
    select
        p.*
        sum(case when cd = 'STG' then 0 else 1 end) 
            over(partition by id order by tmsp) grp
    from #postchg_rows p
) p
where cd = 'STG'
group by id, grp

间隙和孤岛 - 如何按 ID 对每组连续行求和

Gaps and Islands - How to Sum Each Group of Consecutive Rows by ID

sql

sql-server

datetime

window-functions

gaps-and-islands