Question

我们假设我们有一个维护表

Customer LastLogin ActionType
1        12/1/2007 2
1        12/2/2007 2
etc.

我们希望列出所有客户，这些客户在某一年内的任何时间点都有一个或多个不间断的序列，长达14天，以行动类型2登录。

我当然可以使用代码轻松完成此操作，甚至可以在小集合上快速完成。是否有一种非游标方式在SQL中执行它？

Answer 1

这将选择至少两个相同类型的连续操作的所有客户。

WITH    rows AS 
        (
        SELECT  customer, action,
                ROW_NUMBER() OVER (PARTITION BY customer ORDER BY lastlogin) AS rn
        FROM    mytable
        )
SELECT  DISTINCT customer
FROM    rows rp
WHERE   EXISTS
        (
        SELECT  NULL
        FROM    rows rl
        WHERE   rl.customer = rp.customer
                AND rl.rn = rp.rn + 1
                AND rl.action = rp.action
        )

以下是仅针对操作2的更有效查询：

WITH    rows AS 
        (
        SELECT  customer, ROW_NUMBER() OVER (PARTITION BY customer ORDER BY lastlogin) AS rn
        FROM    mytable
        WHERE   action = 2
        )
SELECT  DISTINCT customer
FROM    rows rp
WHERE   EXISTS
        (
        SELECT  NULL
        FROM    rows rl
        WHERE   rl.customer = rp.customer
                AND rl.rn = rp.rn + 1
        )

更新2：

选择不间断的范围：

WITH    rows AS 
        (
        SELECT  customer, action, lastlogin
                ROW_NUMBER() OVER (PARTITION BY customer ORDER BY lastlogin) AS rn
                ROW_NUMBER() OVER (PARTITION BY customer, action ORDER BY lastlogin) AS series
        FROM    mytable
        )
SELECT  DISTINCT customer
FROM    (
        SELECT  customer
        FROM    rows rp
        WHERE   action
        GROUP BY
                customer, actioncode, series - rn
        HAVING
                DETEDIFF(day, MIN(lastlogin), MAX(lastlogin)) >= 14
        ) q

此查询计算两个系列：一个返回连续ORDER BY lastlogin，第二个系列另外按action分区：

action  logindate rn  series diff = rn - series
1       Jan 01    1   1      0
1       Jan 02    2   2      0
2       Jan 03    3   1      2
2       Jan 04    4   2      2
1       Jan 05    5   3      2
1       Jan 06    6   4      2

只要两种方案之间的差异相同，该系列就不会中断。每次中断都会打破这个系列。

这意味着（action, diff）的组合定义了不间断的组。

我们可以按action, diff分组，在群组中找到MAX和MIN并对其进行过滤。

如果您需要选择14行而不是14个连续日，只需过滤COUNT(*)而不是DATEDIFF。

Answer 2

编辑：这对原来的问题大约有两个问题。 14连续是一个不同的答案

首先你需要一个序列，所以你将使用ROWNUMBER

你可以使用ROWNUMBER = ROWNUMBER + 1

为自己做一个自我加入的维护

具有相同客户ID的任何两个具有连续性的行，以及具有“2”ActionType的两行将为您提供CUSTOMER列表作为您的答案。

试试这个

WITH Maintenance AS
(
SELECT 1 as Customer, CONVERT (DateTime, '1/1/2008') DateTimeStamp, 1 ActionType
UNION
SELECT 1, '3/1/2009', 1
UNION
SELECT 1, '3/1/2006', 2
UNION
SELECT 2, '3/1/2009', 1
UNION
SELECT 2, '3/1/2006', 2
)
,RowNumberMaintenance AS
(SELECT  ROW_NUMBER () OVER (ORDER BY Customer, DateTimeStamp)  AS RowNumber, *
FROM Maintenance)
SELECT m1.Customer
From RowNumberMaintenance M1
    INNER JOIN RowNumberMaintenance M2
        ON M1.Customer = M2.Customer
        AND M1.RowNumber = M2.RowNumber + 1
WHERE 1=1
        AND M1.ActionType <> 2
        AND M2.ActionType <> 2

Answer 3

使用：

WITH dates AS (
  SELECT CAST('2007-01-01' AS DATETIME) 'date'
  UNION ALL
   SELECT DATEADD(dd, 1, t.date) 
     FROM dates t
    WHERE DATEADD(dd, 1, t.date) <= GETDATE())
   SELECT m.customer, 
          m.actiontype
     FROM dates d
LEFT JOIN MAINTENANCE m ON m.last_login = d.date
    WHERE m.last_login IS NULL

Answer 4

select customerID, count(customerID)
from maintenance
where actiontype = 2
group by customerID
having count(customerID) >= 1

Answer 5

我将假设一个序列表示两行或更多行具有连续的日期时间值，而对于具有不同操作类型的同一用户，其间没有其他行。既然如此，这应该会给你你想要的东西：

SELECT DISTINCT
     T1.customer
FROM
     Maintenance T1
INNER JOIN Maintenance T2 ON
     T2.customer = T1.customer AND
     T2.action_type = 2 AND
     T2.last_login > T1.last_login
LEFT OUTER JOIN Maintenance T3 ON
     T3.customer = T1.customer AND
     T3.last_login > T1.last_login AND
     T3.last_login < T2.last_login AND
     T3.action_type <> 2
WHERE
     T1.actiontype = 2 AND
     T3.customer IS NULL

SQL就是我上面所说的 - 在它（T2）之后的另一行找到一行（T1），其中action_type = 2，其中没有行（T3）与不同的动作类型。 T3.customer IS NULL检查NULL，因为如果列为NULL（我假设它是一个NOT NULL列），那么这意味着LEFT OUTER JOIN必须找不到符合条件的行。

SQL Server，找到任意值的序列

5 个答案: