Question

我有一个表有这个结构

EmployeeID | LogDate | LogTime | TerminalID | InOut | Accepted

此表处理员工的所有出席

InOut Column = 0 - ＆gt;在（员工刚刚开始他的   移）

Inout Column = 1 - ＆gt; Out（员工刚刚开始   结束他的转变）

我做出选择时的日期样本

例如，上一张图片的EmployeeID = 1009 在2015-10-14（06:56:28）和（16:13:51）有2拳我需要在结果中给我他在这两次2次之间参加的时间这是（09:17:23）

我想要这个结果结构

EmployeeID | LogDate | NumberOfPunshings | State | HoursAttending

我试过这个查询

       select
Employeeid as EmpID,
logdate as logdt,
count(accepted) NumberOfPunshings,
case    count(accepted)                    WHEN 1 THEN 'No Out Punch'
                                           WHEN 2 THEN 'Perfect'
                                           Else 'Null'
                                           End as State,
CONVERT(varchar(12),
       DATEADD(minute,
       DATEDIFF(minute,
       (select top 1 LogTime from Accesslog where Inout=0),
       (select top 1 LogTime from Accesslog where InOut=1)), 0), 114) as HoursAttending

from Accesslog
where Logdate = CAST('2015-10-14' AS datetime)
group by
employeeid,logdate
order by Employeeid

但它不能正常工作因为我想在HoursAttending中只取一个First值（表）

我创建了SQL小提示，向您展示它看起来如何的数据

http://sqlfiddle.com/#!3/0152f9/1

EmpID |           logdt           |  NumberOfPunshings   |  State   |  HoursAttending

1009  | October, 14 2015 00:00:00 |          2           | Perfect  |   08:00:00:000
1088  | October, 15 2015 00:00:00 |          2           | Perfect  |   08:00:00:000

预期结果必须

EmpID |           logdt           |  NumberOfPunshings   |  State   |  HoursAttending

1009  | October, 14 2015 00:00:00 |          2           | Perfect  |   08:00:00:000
1088  | October, 15 2015 00:00:00 |          2           | Perfect  |   06:00:00:000

这些表格在HoursAttending Column中的差异，因为

第一个员工（1009）从08:00开始工作到16:00所以   持续时间是（08:00）正确。

第二名员工（1088）从009：00到15:00开始工作   持续时间必须是（06:00）而不是（08:00）正确。

Answer 1

想象一下，如果我们可以轻松获得数据，那将是可能的，并且我们清楚地了解预期的最终结果应该是什么样的。

SQL Fiddle

MS SQL Server 2014架构设置：

CREATE TABLE Accesslog
    ([EmployeeID] int, [LogDate] datetime, [LogTime] datetime, [Terminal] varchar(11), [InOut] int, [Accepted] int)
;

INSERT INTO Accesslog
    ([EmployeeID], [LogDate], [LogTime], [Terminal], [InOut], [Accepted])
VALUES
    (1009, '2015-10-14 00:00:00', '1900-01-01 06:36:06', 'abcdedghijk', 0, 1),
    (1009, '2015-10-14 00:00:00', '1900-01-01 16:22:41', 'abcdedghijk', 1, 1)
;

查询1 ：

select
        EmployeeID
      , LogDate
      , MIN(case when InOut = 0 then cast(LogTime as time) end) as LogTimeMin
      , MAX(case when InOut = 1 then cast(LogTime as time) end) as LogTimeMax
      , COUNT(*) as CountPunches
      , CONVERT(varchar(12),
         DATEADD(minute,
         DATEDIFF(minute,
         MIN(case when InOut = 0 then cast(LogTime as time) end),
         MAX(case when InOut = 1 then cast(LogTime as time) end)), 0), 114) as HoursAttending
from Accesslog
group by
        EmployeeID
      , LogDate

<强> Results ：

| EmployeeID |          LogDate |  LogTimeMin |  LogTimeMax | CountPunches | HoursAttending |
|------------|------------------|-------------|-------------|--------------|----------------|
|       1009 | October, 14 2015 | 06:36:06.00 | 16:22:41.00 |            2 |   09:46:00:000 |

第2部分

要解决轮班问题，请尝试以下方法。请注意，如果您每天只想要一行，请按查询使用组，但将下面的查询显示为＆＃34;派生表＆＃34; （子查询）

SQL Fiddle

MS SQL Server 2014架构设置：

CREATE TABLE Accesslog
    ([EmployeeID] int, [LogDate] datetime, [LogTime] datetime, [Terminal] varchar(11), [InOut] int, [Accepted] int)
;

INSERT INTO Accesslog
    ([EmployeeID], [LogDate], [LogTime], [Terminal], [InOut], [Accepted])
VALUES
    (1209, '2015-10-14 00:00:00', '1900-01-01 07:00:00', 'abcdedghijk', 0, 1),
    (1209, '2015-10-14 00:00:00', '1900-01-01 12:01:00', 'abcdedghijk', 1, 1),
    (1209, '2015-10-14 00:00:00', '1900-01-01 15:00:00', 'abcdedghijk', 0, 1),
    (1209, '2015-10-14 00:00:00', '1900-01-01 20:02:00', 'abcdedghijk', 1, 1),
    (1009, '2015-10-14 00:00:00', '1900-01-01 08:00:00', 'abcdedghijk', 0, 1),
    (1009, '2015-10-14 00:00:00', '1900-01-01 16:00:00', 'abcdedghijk', 1, 1),
    (1088, '2015-10-15 00:00:00', '1900-01-01 09:00:00', 'aaaa', 0, 1),
    (1088, '2015-10-15 00:00:00', '1900-01-01 15:00:00', 'aaaa', 1, 1)
;

查询1 ：

/*
 including (07:00 - 12:00) shift two (15:00 - 20:00) 
 */

WITH CTE as (
        SELECT *, ROW_NUMBER() OVER(PARTITION BY EmployeeID, LogDate, InOut
                                    ORDER BY LogTime ASC) AS shiftno
        FROM Accesslog
             )
SELECT
        ins.EmployeeID
      , ins.LogDate
      , ins.LogTime  as LogTimeIn
      , outs.LogTime as LogTimeOut
      , ins.Accepted + ISNULL(outs.Accepted,0) as CountPunches
      , CONVERT(varchar(12),
         DATEADD(minute,
         DATEDIFF(minute,
         ins.LogTime,
         outs.LogTime), 0), 114) as HoursAttending
FROM CTE  AS ins
INNER JOIN CTE AS outs ON ins.InOut = 0 AND outs.InOut = 1
                      AND ins.EmployeeID = outs.EmployeeID
                      AND ins.LogDate = outs.LogDate
                      AND ins.shiftno = outs.shiftno

<强> Results ：

| EmployeeID |                   LogDate |                 LogTimeIn |                LogTimeOut | CountPunches | HoursAttending |
|------------|---------------------------|---------------------------|---------------------------|--------------|----------------|
|       1009 | October, 14 2015 00:00:00 | January, 01 1900 08:00:00 | January, 01 1900 16:00:00 |            2 |   08:00:00:000 |
|       1088 | October, 15 2015 00:00:00 | January, 01 1900 09:00:00 | January, 01 1900 15:00:00 |            2 |   06:00:00:000 |
|       1209 | October, 14 2015 00:00:00 | January, 01 1900 07:00:00 | January, 01 1900 12:01:00 |            2 |   05:01:00:000 |
|       1209 | October, 14 2015 00:00:00 | January, 01 1900 15:00:00 | January, 01 1900 20:02:00 |            2 |   05:02:00:000 |

SQL Server如何在组内创建组 - 使用GROUP BY

1 个答案: