刪除從表中共享相同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 | 

我想從一個登錄名和共享相同ID的註銷之間的表都刪除，但保留一切那裏。 somedate字段是日期時間。表中可能會有更多登錄/註銷，並且會有登錄沒有相應的註銷。我想讓它保持在那裏，因爲註銷最終會出現。

我正在考慮使用遊標。哪一個對性能來說最好呢？決賽桌可能有幾百萬行。

中刪除該表應該是這樣的後：

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

編輯：登錄和註銷，包括登錄/註銷行之間刪除所有內容。

Answer 1

假設每個註銷有刪除一切相應的登錄，你可以試試這個：

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

其結果將是：

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

如果你想與 「登錄」 和 「註銷」 行過，你可以這樣做：

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

你會得到這樣的結果：

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問題，編輯更正：

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') 

以上應該從你的表，其中登錄有相應的註銷

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

有是一些不明確的情況（如果有沒有匹配登錄的命令會發生什麼？如果登錄與登出同一天，我們假設登出後登錄？等等），但這應該給你一個出發點;

它刪除它們不登錄或註銷記錄的所有記錄，其中記錄有較早登錄，以及相同ID的後註銷記錄;

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

讓我們來做一些測試數據;

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') 

我會使用這樣的查詢。已連接的查詢將返回登錄和註銷日期時間之間的特定ID的所有行。它不會刪除ID 2的任何內容，因爲還沒有註銷。

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 

後，這樣的結果已經運行將是

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 

編輯：更新我的回答，以便它現在刪除登錄/註銷命令了。