我有一张这样的桌子：
customer_id mis_date status
----------------------------
10003 2014-01-01 1
10003 2014-01-02 1
10003 2014-01-03 0
10003 2014-01-04 0
10003 2014-01-05 0
10003 2014-01-06 1
10003 2014-01-07 1
10003 2014-01-08 1
10003 2014-01-09 1
10003 2014-01-10 0
10003 2014-01-11 0
10003 2014-01-12 0
10003 2014-01-13 1
10003 2014-01-14 1
10003 2014-01-15 1
我正在尝试建立“组”列：
customer_id mis_date status group
----------------------------------
10003 2014-01-01 1 1
10003 2014-01-02 1 1
10003 2014-01-03 0 NULL
10003 2014-01-04 0 NULL
10003 2014-01-05 0 NULL
10003 2014-01-06 1 2
10003 2014-01-07 1 2
10003 2014-01-08 1 2
10003 2014-01-09 1 2
10003 2014-01-10 0 NULL
10003 2014-01-11 0 NULL
10003 2014-01-12 0 NULL
10003 2014-01-13 1 3
10003 2014-01-14 1 3
10003 2014-01-15 1 3
有人知道我如何构建此组列吗？
逻辑：我每天都在跟踪客户状态，并且我想每天都知道该状态在客户历史记录中发生过多少次，但只有当他处于状态时才发生。
例如：first_time-1，second_time-2等等
我踢开了头，找不到解决方法。我想这不是那么复杂。
谢谢！
答案 0 :(得分：3)
类似的事情应该起作用：
;WITH CTE AS (
SELECT customer_id, mis_date, status,
ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY mis_date) -
ROW_NUMBER() OVER (PARTITION BY customer_id, status ORDER BY mis_date) AS grp
FROM mytable
), CTE2 AS (
SELECT customer_id, status, grp,
ROW_NUMBER() OVER (ORDER BY MIN(mis_date)) AS rn
FROM CTE
WHERE status = 1
GROUP BY customer_id, status, grp
)
SELECT c.customer_id, c.mis_date, c.status, rn
FROM CTE c
LEFT JOIN CTE2 c2
ON c.customer_id = c2.customer_id AND c.status = c2.status AND c.grp = c2.grp
ORDER BY mis_date
CTE标识具有相同status值的连续记录的孤岛。 CTE2枚举status = 1个子组。
答案 1 :(得分：3)
没有CTE的另一种方法类似于以下查询。
SELECT customer_id, mis_date, status,
CASE WHEN status = 0 THEN NULL ELSE Dense_rank() OVER (ORDER BY rc) END grp
FROM (SELECT *,
(SELECT CASE WHEN status = 0 THEN 0
ELSE (SELECT Count(status) FROM table1 t2
WHERE t2.mis_date <= t1.mis_date AND status = 0) END grp)rc
FROM table1 t1) t2
ORDER BY mis_date
输出：
+-------------+-------------------------+--------+------+
| customer_id | mis_date | status | grp |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-01 00:00:00.000 | 1 | 1 |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-02 00:00:00.000 | 1 | 1 |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-03 00:00:00.000 | 0 | NULL |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-04 00:00:00.000 | 0 | NULL |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-05 00:00:00.000 | 0 | NULL |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-06 00:00:00.000 | 1 | 2 |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-07 00:00:00.000 | 1 | 2 |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-08 00:00:00.000 | 1 | 2 |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-09 00:00:00.000 | 1 | 2 |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-10 00:00:00.000 | 0 | NULL |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-11 00:00:00.000 | 0 | NULL |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-12 00:00:00.000 | 0 | NULL |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-13 00:00:00.000 | 1 | 3 |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-14 00:00:00.000 | 1 | 3 |
+-------------+-------------------------+--------+------+
| 10003 | 2014-01-15 00:00:00.000 | 1 | 3 |
+-------------+-------------------------+--------+------+
答案 2 :(得分：2)
请检查此解决方案。这会根据您的需要添加分组
with cte0 as
(
select [customer_id], [mis_date], [status],
COALESCE(LAG(status) over (order by mis_date), status) oldstatus
FRom Table1
),
cte1 as (
select cte0.*,
case when status = 0 then
null
else
COUNT( case when status != oldStatus and status = 0 then 1 else null end) OVER (ORDER BY mis_date)
end + 1 grp
from cte0
)
select * from cte1
GO
customer_id | mis_date | status | oldstatus | grp
----------: | :------------------ | -----: | --------: | ---:
10003 | 01/01/2014 00:00:00 | 1 | 1 | 1
10003 | 02/01/2014 00:00:00 | 1 | 1 | 1
10003 | 03/01/2014 00:00:00 | 0 | 1 | null
10003 | 04/01/2014 00:00:00 | 0 | 0 | null
10003 | 05/01/2014 00:00:00 | 0 | 0 | null
10003 | 06/01/2014 00:00:00 | 1 | 0 | 2
10003 | 07/01/2014 00:00:00 | 1 | 1 | 2
10003 | 08/01/2014 00:00:00 | 1 | 1 | 2
10003 | 09/01/2014 00:00:00 | 1 | 1 | 2
10003 | 10/01/2014 00:00:00 | 0 | 1 | null
10003 | 11/01/2014 00:00:00 | 0 | 0 | null
10003 | 12/01/2014 00:00:00 | 0 | 0 | null
10003 | 13/01/2014 00:00:00 | 1 | 0 | 3
10003 | 14/01/2014 00:00:00 | 1 | 1 | 3
10003 | 15/01/2014 00:00:00 | 1 | 1 | 3
答案 3 :(得分：1)
您可以通过非零状态之前的数量来识别每组“ 1”。如果您不关心组号是连续的：
select t.*,
(case when status = 1
then sum(case when status = 0 then 1 else 0 end) over (partition by customer_id order by mis_date)
end) as grp
from t;
没有子查询，联接或聚集。
但是，您可能希望数字是连续的（如您的示例所示）。为此，需要一个子查询：
select t.*,
(case when status = 1
then dense_rank() over (partition by customer_id order by grp1)
end) as grp
from (select t.*,
sum(case when status = 0 then 1 else 0 end) over (partition by customer_id order by mis_date) as grp1
from t
) t
答案 4 :(得分：0)
您可以在SQL Server中使用ALTER TABLE语句向表中添加列。 语法
在SQL Server（Transact-SQL）的表中添加列的语法为：
ALTER TABLE table_name
ADD column_name column_definition;
让我们看一个示例，该示例显示如何使用ALTER TABLE语句在SQL Server表中添加列。
例如：
ALTER TABLE customer
ADD group VARCHAR(10);
此SQL Server ALTER TABLE示例将在客户表中添加一列，称为group。