计算SQL中连续分组项的数目

我想使用SQL（SQL server）在下面看到的“当前状态”字段中创建并填充以下条目数

有没有一种快速的方法可以使用类似于

row\u number（）

的东西来创建我要查找的字段？单靠这一点似乎是不够的

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谢谢

如果序列是标识列，则可以执行以下操作：

select t.*, 
       row_number() over (partition by (Sequence - seq) order by Sequence) as [No. of Entries in Curr.Status]
from (select t.*, 
             row_number() over (partition by [Curr.Status] order by Sequence) as seq
      from table t
     ) t;

否则，您需要生成两个

行编号

：

select t.*, 
       row_number() over (partition by (seq1- seq2) order by Sequence) as [No. of Entries in Curr.Status]
from (select t.*, 
             row_number() over (partition by id order by Sequence) as seq1
             row_number() over (partition by id, [Curr.Status] order by Sequence) as seq2
      from table t
     ) t;

---

这似乎是一个群体和岛屿问题。然而，关于如何实现这一点，有很多例子：

WITH VTE AS(
        SELECT *
        FROM (VALUES('9-9999-9',1 ,'Status D','Status A'),
                    ('9-9999-9',2 ,'Status A','Status A'),
                    ('9-9999-9',3 ,'Status A','Status A'),
                    ('9-9999-9',4 ,'Status A','Status A'),
                    ('9-9999-9',5 ,'Status A','Status B'),
                    ('9-9999-9',6 ,'Status B','Status B'),
                    ('9-9999-9',7 ,'Status B','Status B'),
                    ('9-9999-9',8 ,'Status B','Status A'),
                    ('9-9999-9',9 ,'Status A','Status A'),
                    ('9-9999-9',10,'Status A','Status C'),
                    ('9-9999-9',11,'Status C','Status C')) V(ID, Sequence, PrevStatus,CurrStatus)),
CTE AS(            
    SELECT ID,
           [Sequence],
           PrevStatus,
           CurrStatus,
           ROW_NUMBER() OVER (PARTITION BY ID ORDER BY [Sequence]) -
           ROW_NUMBER() OVER (PARTITION BY ID,CurrStatus ORDER BY [Sequence]) AS Grp
    FROM VTE V)
SELECT ID,
       [Sequence],
       PrevStatus,
       CurrStatus,
       ROW_NUMBER() OVER (PARTITION BY Grp ORDER BY [Sequence]) AS Entries
FROM CTE;

---

您可以使用

LAG

函数标记状态更改的行，并使用

SUM（）OVER（）

为每个组分配唯一的编号。组内的编号很简单：

DECLARE @t TABLE (ID VARCHAR(100), Sequence INT, PrevStatus VARCHAR(100), CurrStatus VARCHAR(100));
INSERT INTO @t VALUES
('9-9999-9',  1, 'Status D', 'Status A'),
('9-9999-9',  2, 'Status A', 'Status A'),
('9-9999-9',  3, 'Status A', 'Status A'),
('9-9999-9',  4, 'Status A', 'Status A'),
('9-9999-9',  5, 'Status A', 'Status B'),
('9-9999-9',  6, 'Status B', 'Status B'),
('9-9999-9',  7, 'Status B', 'Status B'),
('9-9999-9',  8, 'Status B', 'Status A'),
('9-9999-9',  9, 'Status A', 'Status A'),
('9-9999-9', 10, 'Status A', 'Status C'),
('9-9999-9', 11, 'Status C', 'Status C');

WITH cte1 AS (
    SELECT *, CASE WHEN LAG(CurrStatus) OVER(ORDER BY Sequence) = CurrStatus THEN 0 ELSE 1 END AS chg
    FROM @t
), cte2 AS (
    SELECT *, SUM(chg) OVER(ORDER BY Sequence) AS grp
    FROM cte1
), cte3 AS (
    SELECT *, ROW_NUMBER() OVER(PARTITION BY grp ORDER BY Sequence) AS SeqInGroup
    FROM cte2
)
SELECT *
FROM cte3
ORDER BY Sequence

---

您的尝试是什么？请分享。这个数字是在什么领域计算的？谢谢。通过一些快速的改变，这解决了我的问题。知道“r1-r2分区”可以用于此类操作非常有帮助！

DECLARE @t TABLE (ID VARCHAR(100), Sequence INT, PrevStatus VARCHAR(100), CurrStatus VARCHAR(100));
INSERT INTO @t VALUES
('9-9999-9',  1, 'Status D', 'Status A'),
('9-9999-9',  2, 'Status A', 'Status A'),
('9-9999-9',  3, 'Status A', 'Status A'),
('9-9999-9',  4, 'Status A', 'Status A'),
('9-9999-9',  5, 'Status A', 'Status B'),
('9-9999-9',  6, 'Status B', 'Status B'),
('9-9999-9',  7, 'Status B', 'Status B'),
('9-9999-9',  8, 'Status B', 'Status A'),
('9-9999-9',  9, 'Status A', 'Status A'),
('9-9999-9', 10, 'Status A', 'Status C'),
('9-9999-9', 11, 'Status C', 'Status C');

WITH cte1 AS (
    SELECT *, CASE WHEN LAG(CurrStatus) OVER(ORDER BY Sequence) = CurrStatus THEN 0 ELSE 1 END AS chg
    FROM @t
), cte2 AS (
    SELECT *, SUM(chg) OVER(ORDER BY Sequence) AS grp
    FROM cte1
), cte3 AS (
    SELECT *, ROW_NUMBER() OVER(PARTITION BY grp ORDER BY Sequence) AS SeqInGroup
    FROM cte2
)
SELECT *
FROM cte3
ORDER BY Sequence