删除从表中共享相同ID的所有行

Question

我有如下表：

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 | 

我想从一个登录名和共享相同ID的注销之间的表都删除，但保留一切那里。 somedate字段是日期时间。表中可能会有更多登录/注销，并且会有登录没有相应的注销。我想让它保持在那里，因为注销最终会出现。

我正在考虑使用游标。哪一个对性能来说最好呢？决赛桌可能有几百万行。

中删除该表应该是这样的后：

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

编辑：登录和注销，包括登录/注销行之间删除所有内容。

Answer 1

假设每个注销有删除一切相应的登录，你可以试试这个：

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

其结果将是：

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

如果你想与 “登录” 和 “注销” 行过，你可以这样做：

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 2

我想了解我的第一篇文章你的问题，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') 

以上应该从你的表，其中登录有相应的注销

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

有是一些不明确的情况（如果有没有匹配登录的命令会发生什么？如果登录与登出同一天，我们假设登出后登录？等等），但这应该给你一个出发点;

它删除它们不登录或注销记录的所有记录，其中记录有较早登录，以及相同ID的后注销记录;

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

让我们来做一些测试数据;

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。已连接的查询将返回登录和注销日期时间之间的特定ID的所有行。它不会删除ID 2的任何内容，因为还没有注销。

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 

后，这样的结果已经运行将是

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：更新我的回答，以便它现在删除登录/注销命令了。