2016-01-23 87 views
3

我登录的表看起来像这样:选择列成列分组行的值

登录表

Emp_ID Created    | Action 
1  20/01/2016 10:44:42 AM  login 
1  20/01/2016 4:45:49 PM  logout 
1  20/01/2016 6:30:13 PM  logout 
1  21/01/2016 8:46:28 AM  login 
1  21/01/2016 9:46:42 AM  login 
1  21/01/2016 1:46:46 PM  logout 
1  22/01/2016 8:49:21 AM  login 
1  22/01/2016 1:49:27 PM  logout 
1  22/01/2016 2:29:53 PM  login 
1  22/01/2016 2:30:13 PM  logout 
3  22/01/2016 2:42:06 PM  login 
1  22/01/2016 9:57:22 PM  login 
1  22/01/2016 10:22:23 PM  logout 
1  23/01/2016 8:01:47 AM  login 
1  23/01/2016 9:01:58 AM  logout 
3  23/01/2016 8:02:06 AM  login 
3  23/01/2016 9:02:28 AM  logout 

员工表

| ID | Fname | Lname | 
|----|-------|-------| 
| 1 | James | Brown | 
| 2 | Mark | Bond | 
| 3 | Kemi | Ojo | 

结果
| created | login | logout | Employee | Emp_ID | 
|------------|----------|----------|-------------|--------| 
| 2016-01-20 | 10:44:42 | 18:30:13 | James Brown |  1 | 
| 2016-01-21 | 08:46:28 | 13:46:46 | James Brown |  1 | 
| 2016-01-22 | 08:49:21 | 22:22:23 | James Brown |  1 | 
| 2016-01-22 | 14:42:06 | 22:22:23 | Kemi Ojo |  3 | 
| 2016-01-23 | 08:01:47 | 09:02:28 | James Brown |  1 | 
| 2016-01-23 | 08:02:06 | 09:02:28 | Kemi Ojo |  3 | 

这是我曾尝试过:

SELECT 
    CAST(LI.created AS DATE) AS created, 
    MIN(CAST(LI.created AS TIME)) AS login, 
    MAX(CAST(LO.created AS TIME)) AS logout, 
    e.fname+' '+e.lname Employee, li.Emp_ID 

FROM 
    Logins LI 
LEFT OUTER JOIN Logins LO ON 
    LO.action = 'logout' AND 
    CAST(LO.created AS DATE) = CAST(LI.created AS DATE) 
    JOIN dbo.Employees AS E ON E.ID = li.Emp_ID 

WHERE 
    LI.action = 'login' 
GROUP BY 
    CAST(LI.created AS DATE), E.fname + ' ' + E.lname, li.Emp_ID 

但是结果不正确。

  1. 请注意,不同用户的最后两个结果是相同的。例如09:02:28出现两次而不是9:01:58

  2. 另外我有一个登录时没有注销emp_id = 3的问题。这发生在应用程序意外关闭时。

3我如何可以放置一个在当没有注销 4.或者,你们会在这种情况下该怎么做的建议的情况下00:00:00?

我需要选择的结果集看起来像这样:

| created | login | logout | Employee | Emp_ID | 
|------------|----------|----------|-------------|--------| 
| 2016-01-20 | 10:44:42 | 18:30:13 | James Brown |  1 | 
| 2016-01-21 | 08:46:28 | 13:46:46 | James Brown |  1 | 
| 2016-01-22 | 08:49:21 | 22:22:23 | James Brown |  1 | 
| 2016-01-22 | 14:42:06 | 00:00:00 | Kemi Ojo |  3 | 
| 2016-01-23 | 08:01:47 | 09:01:58 | James Brown |  1 | 
| 2016-01-23 | 08:02:06 | 09:02:28 | Kemi Ojo |  3 | 

SQL fiddle

+1

这是一个非常好的问题:清晰,示例数据,您自己的尝试,错误的输出,预期的输出。 Thx为此,我的身边+1! – Shnugo

+1

是的。这个问题的链接应该在'如何提问问题'页面中专门针对SQL问题给出。 – Utsav

回答

3

首先,一个知情问题的赞美。表结构,小提琴演示和预期输出帮助很大。

现在我试过了,它在小提琴中工作。请重新检查并让我知道。

select t_login.emp_id,t_login.dt_created as created,t_login.login, 
case when t_logout.logout is null 
    then cast('00:00:00' as time) 
else t_logout.logout end as logout, 
e.fname+' '+e.lname Employee 
from 
    (select emp_id,CAST(created AS DATE) AS dt_created, 
    MIN(CAST(created AS TIME)) as login 
    from logins 
    where action='login' 
    group by emp_id,CAST(created AS DATE)) t_login 
left join 
    (select emp_id,CAST(created AS DATE) AS dt_created, 
    max(CAST(created AS TIME)) as logout 
    from logins 
    where action='logout' 
    group by emp_id,CAST(created AS DATE)) t_logout 
     on t_login.emp_id=t_logout.emp_id 
     and t_login.dt_created=t_logout.dt_created 
inner join 
    employees e 
     on e.id=t_login.emp_id 

PS:这不会处理在特定日期没有登录并且只有注销的情况。如果您需要,则使用full outer join并使用与select条款中使用的case相同的语句。

看到这里

http://sqlfiddle.com/#!3/465f0/34

小提琴演示输出继电器

+---------+-------------+-------------------+-------------------+-------------+ 
| emp_id | created |  login  |  logout  | Employee | 
+---------+-------------+-------------------+-------------------+-------------+ 
|  1 | 2016-01-20 | 10:44:42.0000000 | 18:30:13.0000000 | James Brown | 
|  1 | 2016-01-21 | 08:46:28.0000000 | 13:46:46.0000000 | James Brown | 
|  1 | 2016-01-22 | 08:49:21.0000000 | 22:22:23.0000000 | James Brown | 
|  3 | 2016-01-22 | 14:42:06.0000000 | 00:00:00.0000000 | Kemi Ojo | 
|  1 | 2016-01-23 | 08:01:47.0000000 | 09:01:58.0000000 | James Brown | 
|  3 | 2016-01-23 | 08:02:06.0000000 | 09:02:28.0000000 | Kemi Ojo | 
+---------+-------------+-------------------+-------------------+-------------+ 
+1

好的解决方案,它在技术上与我的方法相同,区别在于,当我将它们打包在CTE中时,使用了嵌套的子选择。 +1从我身边 – Shnugo

+0

我建议编辑摆脱外部SELECT中的CASE WHEN。 :) –

+1

嗨@DavidBachmannJeppesen - 欣赏你建议编辑的努力。 'case'声明适用于所有RDBMS。但是我不确定'isnull'。请不要介意。无论如何感谢。祝你有美好的一天:) – Utsav

2

你错过了LO表和李表之间的连接:

AND LO.Emp_ID = LI.Emp_ID 

我认为你可以做到这一点,而不使用自联接,而是使用CASE WHEN,我认为这会更漂亮,但试试这个补充。从评论

编辑: 只需更换: MAX(CAST(LO.Created随着时间)) ...有: MAX(CAST(ISNULL(LO.Created,'00:00:00' )AS时间))

+0

这项工作,但没有照顾在当天没有注销的日期登录 – Smith

3

这是我的建议(注意文化的具体日期文字的,我必须设置一个语言获得正确的日期转换):

SET LANGUAGE GERMAN; 

DECLARE @logins TABLE(Emp_ID INT,Created DATETIME, Action VARCHAR(100)); 
INSERT INTO @logins VALUES 
(1,'20/01/2016 10:44:42 AM','login') 
,(1,'20/01/2016 4:45:49 PM','logout') 
,(1,'20/01/2016 6:30:13 PM','logout') 
,(1,'21/01/2016 8:46:28 AM','login') 
,(1,'21/01/2016 9:46:42 AM','login') 
,(1,'21/01/2016 1:46:46 PM','logout') 
,(1,'22/01/2016 8:49:21 AM','login') 
,(1,'22/01/2016 1:49:27 PM','logout') 
,(1,'22/01/2016 2:29:53 PM','login') 
,(1,'22/01/2016 2:30:13 PM','logout') 
,(3,'22/01/2016 2:42:06 PM','login') 
,(1,'22/01/2016 9:57:22 PM','login') 
,(1,'22/01/2016 10:22:23 PM','logout') 
,(1,'23/01/2016 8:01:47 AM','login') 
,(1,'23/01/2016 9:01:58 AM','logout') 
,(3,'23/01/2016 8:02:06 AM','login') 
,(3,'23/01/2016 9:02:28 AM','logout'); 

DECLARE @employees TABLE(ID INT,Fname VARCHAR(100),Lname VARCHAR(100)); 
INSERT INTO @employees VALUES 
(1,'James','Brown') 
,(2,'Mark','Bond') 
,(3,'Kemi','Ojo'); 

WITH Logins AS 
(
    SELECT 
     MIN(PureDate) AS Created 
     ,MIN(CAST(l.Created AS TIME)) AS LoginTime 
     ,l.Emp_ID 
    FROM @logins AS l 
    CROSS APPLY(SELECT CAST(l.Created AS DATE) PureDate) AS Created 
    WHERE l.Action ='login' 
    GROUP BY l.Emp_ID,PureDate 
) 
,Logouts AS 
(
    SELECT 
     MAX(PureDate) AS Created 
     ,MAX(CAST(l.Created AS TIME)) AS LogoutTime 
     ,l.Emp_ID 
    FROM @logins AS l 
    CROSS APPLY(SELECT CAST(l.Created AS DATE) PureDate) AS Created 
    WHERE l.Action ='logout' 
    GROUP BY l.Emp_ID,PureDate 
) 
SELECT Logins.Created 
     ,Logins.LoginTime 
     ,ISNULL(Logouts.LogoutTime,'00:00:00') 
     ,e.Lname 
     ,e.ID 
FROM Logins 
INNER JOIN @employees AS e ON e.ID = Logins.Emp_ID 
LEFT JOIN Logouts ON Logins.Emp_ID = Logouts.Emp_ID 
       AND Logins.Created = Logouts.Created 
ORDER BY Created,LoginTime 

结果

2016-01-20 10:44:42 18:30:13 Brown 1 
2016-01-21 08:46:28 13:46:46 Brown 1 
2016-01-22 08:49:21 22:22:23 Brown 1 
2016-01-22 14:42:06 00:00:00 Ojo  3 
2016-01-23 08:01:47 09:01:58 Brown 1 
2016-01-23 08:02:06 09:02:28 Ojo  3 
+0

我可以在小提琴中测试你的代码,但不得不花费时间让它在代码中工作,我还没有得到 – Smith

+0

@Smith要测试这个非常简单:只需将整个代码粘贴到一个空的查询窗口并执行。这与声明的表变量一起工作(它们以'@'开头),因此您不需要任何现有的结构来测试它......使用它只需要将表变量更改为想要读取的实际表从。 – Shnugo