删除 table 中共享相同 ID 的所有行

Question

我有以下 table:

Name     |   ID  |  date    |
Login    |   1   | somedate |
Command  |   1   | somedate |
Command  |   1   | somedate |
Login    |   2   | somedate |
Command  |   1   | somedate |
Command  |   2   | somedate |
Logout   |   1   | somedate |
Command  |   2   | somedate |

我想从 table 中删除共享相同 ID 的登录和注销之间的所有内容，但保留其他所有内容。 somedate 字段是日期时间。 table 中可能有更多的 logins/logouts，并且会有没有相应注销的登录。我希望它保留在那里，因为注销最终会出现。

我正在考虑使用游标。哪种也是提高性能的最佳方法？最后的 table 可能有几百万行。

删除后 table 应如下所示：

Name    |  ID  |  
Login   |  2   |
Command |  2   |
Command |  2   |

编辑：删除登录和注销之间的所有内容，包括带有 Login/Logout 的行。

Answer 1

第一次 post 时我想不明白你的问题，编辑更正：

delete from yourTable
where id in (select yt.Id from yourtable yt
        inner join yourtable yt2 on yt2.Id = yt.Id and yt2.Name like 'Logout'
        where yt.Name like 'Login')

上面应该删除你的table中的所有内容，其中登录有相应的注销

Answer 2

假设每次注销都有相应的登录，您可以试试这个：

DELETE
FROM yourTable
WHERE ID IN
(
    SELECT ID
    FROM yourTable
    WHERE Name LIKE 'Logout'
)

结果将是：

Name    |  ID  |
Login   |  2   |
Command |  2   |
Command |  2   |

如果您也想要包含 "Login" 和 "Logout" 的行，您可以这样做：

DELETE
FROM yourTable
WHERE ID IN
(
    SELECT ID
    FROM yourTable
    WHERE Name LIKE 'Logout'
)
AND Name NOT LIKE 'Login'
AND Name NOT LIKE 'Logout'

你会得到这样的结果：

Name    |  ID  |
Login   |  1   |
Login   |  2   |
Command |  2   |
Logout  |  1   |
Command |  2   |

Answer 3

select distinct(id)  into @id from your_table where Name = 'Logout'; -- will give you all user id that need to delete
delete from your_table where id in (@id );

Answer 4

有一些模棱两可的情况（如果有没有匹配登录的命令会发生什么？如果登录与注销在同一天，我们是否假设注销是在登录之后？等等）但这应该给你一个起点；

它删除所有非登录或注销记录，其中记录有较早登录，和稍后相同Id的登出记录；

WITH TestData AS (
    SELECT 'Login' as Name, 1 AS ID, cast('01/01/2000' as date) as Date
    UNION ALL
    SELECT 'Command' as Name, 1 AS ID, cast('02/01/2000' as date) as Date
    UNION ALL
    SELECT 'Command' as Name, 1 AS ID, cast('03/01/2000' as date) as Date
    UNION ALL
    SELECT 'Logout' as Name, 1 AS ID, cast('04/01/2000' as date) as Date
    UNION ALL
    SELECT 'Command' as Name, 1 AS ID, cast('05/01/2000' as date) as Date
    UNION ALL
    SELECT 'Command' as Name, 2 AS ID, cast('01/01/2000' as date) as Date
)
DELETE FROM TestData td1
-- Delete any records which are not login or logout
WHERE Name <> 'Login' AND Name <> 'Logout'
-- Where there is an earlier Login
AND EXISTS (SELECT 1 from TestData td2 where td2.Name = 'Login' AND td1.Id = td2.Id AND td2.Date <= td1.Date)
-- and a later logout
AND EXISTS (SELECT 1 from TestData td3 where td3.Name = 'Logout' AND td1.Id = td3.Id AND td3.Date >= td1.Date)

Answer 5

我们做一些测试数据；

IF OBJECT_ID('tempdb..#TestData') IS NOT NULL DROP TABLE #TestData
GO
CREATE TABLE #TestData (Name varchar(7), ID int, Date datetime)
INSERT INTO #TestData (Name, ID, Date)
VALUES
 ('Login',1,'2016-06-23 12:00:00')
,('Command',1,'2016-06-23 12:05:00')
,('Command',1,'2016-06-23 12:10:00')
,('Login',2,'2016-06-23 12:15:00')
,('Command',1,'2016-06-23 12:20:00')
,('Command',2,'2016-06-23 12:25:00')
,('Logout',1,'2016-06-23 12:30:00')
,('Command',2,'2016-06-23 12:35:00')

我会使用这样的查询。连接查询 returns 特定 ID 在登录和注销日期时间之间的所有行。它不会删除 ID 2 的任何内容，因为该 ID 2 还没有注销。

DELETE TD
FROM #TestData TD
INNER JOIN 
(
SELECT a.Name, a.ID, a.Date
FROM #TestData a
LEFT JOIN (
            SELECT ID, Date
            FROM #TestData
            WHERE Name = 'Login'
          ) lin
ON a.ID = lin.ID
LEFT JOIN (
            SELECT ID, Date
            FROM #TestData
            WHERE Name = 'Logout'
          ) lout
ON a.ID = lout.ID
WHERE a.Date BETWEEN lin.Date AND lout.Date
) b
ON TD.Date = b.Date
AND TD.ID = b.ID
AND TD.Name = b.Name

这运行之后的结果将是

Name    ID  Date
Login   2   2016-06-23 12:15:00.000
Command 2   2016-06-23 12:25:00.000
Command 2   2016-06-23 12:35:00.000

编辑：更新了我的答案，现在它也删除了 login/logout 命令。

删除 table 中共享相同 ID 的所有行

Deleting all rows from table that share the same ID

sql

tsql

sap-ase