Sql 如何透视Datetime类型实现基于日期的检查

我想透视“Person\u Log”表数据。。 其栏目如下：

EmployeeID-> Foreign key
Log-> DateTime type

“日志”的格式为“2013年1月22日下午2:02:34”

我想根据日志列中的日期检查创建数据透视，然后显示每个日期时间的最小值和最大值。。。 这是一种考勤报告。。 必需的列如下所示

EmployeeID, 01-Jan IN, 01-Jan OUT, 02-Jan IN, 02-Jan OUT, 03-Jan IN, 03-Jan OUT.....and so on..

EmployeeID以外的列应该只包含从“日志”列提取的时间。。 对于提取，我使用convert（char（10），Log，101）表示日期，使用convert（char（5），Log，108）表示时间提取

我一天所能达到的最好成绩是：

SELECT   dbo.DoorLog.EmployeeID, 
         CONVERT(char(10), 
         MIN(dbo.DoorLog.DateTime), 101) AS Date, 
         CONVERT(char(8), MIN(dbo.DoorLog.DateTime), 108) AS INTime, 
         CONVERT(char(8), MAX(dbo.DoorLog.DateTime), 108) AS OUTTime, 
         dbo.Person.Name, dbo.Person.Department, dbo.Person.Sex, 
         dbo.Person.WorkUnit, 
         dbo.Person.Position
FROM dbo.DoorLog 
INNER JOIN dbo.Person ON dbo.DoorLog.EmployeeID = dbo.Person.EmployeeID
GROUP BY CONVERT(char(10), dbo.DoorLog.DateTime, 101), 
         dbo.DoorLog.EmployeeID, dbo.Person.Name, dbo.Person.Department, 
         dbo.Person.Sex, dbo.Person.WorkUnit, dbo.Person.Position;

请回复，因为我将在两天的截止日期前完成。。 提前谢谢

如你所问…样本数据

Log                           EmployeeID 
2013/01/31 12:31              11
2013/01/25 10:31              10
2013/01/23 13:29              8
2013/01/20 11:49              4

---

此数据转换是一个复杂的过程。在SQL Server 2005+中，有一个函数可以为您旋转数据。有几种方法可以让你得到你想要的结果。两个版本都将实现

UNPIVOT

和

PIVOT

功能

样本数据：

CREATE TABLE Person ([EmployeeId] int, [Name] varchar(4));

INSERT INTO Person  ([EmployeeId], [Name])
VALUES
    (11, 'Jim'),
    (10, 'John'),
    (8, 'Mary'),
    (4, 'Tim');

CREATE TABLE DoorLog([EmployeeId] int, [DoorDate] datetime);

INSERT INTO DoorLog ([EmployeeId], [DoorDate])
VALUES
    (11, '2013-01-31 12:31:00'),
    (11, '2013-01-31 16:50:00'),
    (11, '2013-01-31 17:50:00'),
    (10, '2013-01-25 10:31:00'),
    (10, '2013-01-25 16:45:00'),
    (8, '2013-01-23 13:29:00'),
    (8, '2013-01-23 18:25:00'),
    (4, '2013-01-20 11:49:00'),
    (4, '2013-01-20 19:10:00'),
    (11, '2013-01-15 11:15:00'),
    (11, '2013-01-15 16:25:00'),
    (10, '2013-01-10 09:21:00'),
    (10, '2013-01-10 15:45:00'),
    (8, '2013-01-08 01:29:00'),
    (8, '2013-01-08 02:25:00'),
    (4, '2013-01-06 10:17:00'),
    (4, '2013-01-06 19:10:00');

您的查询首先获取每个日期具有最小/最大值的员工列表：

select p.employeeid,
  p.name,
  convert(char(10),d.doordate, 101) date,
  min(d.doordate) [In],
  max(d.doordate) [Out]
from person p
left join doorlog d
  on p.employeeid = d.employeeid
group by p.employeeid, p.name, 
  convert(char(10),d.doordate, 101)

看

下一步是

UNPIVOT

，它将获取输入/输出时间的单独列，并将它们放入多行：

select employeeid, name, 
      convert(char(8), doortime, 108) DoorTime,
      date + '_'+ col as col_names
from
(
  select p.employeeid,
      p.name,
      convert(char(10),d.doordate, 101) date,
      min(d.doordate) [In],
      max(d.doordate) [Out]
  from person p
  left join doorlog d
      on p.employeeid = d.employeeid
  group by p.employeeid, p.name, 
      convert(char(10),d.doordate, 101)
) src
unpivot
(
  doortime
  for col in ([In], [Out])
) unpiv

看。结果如下所示：

| EMPLOYEEID | NAME | DOORTIME |      COL_NAMES |
-------------------------------------------------
|          4 |  Tim | 10:17:00 |  01/06/2013_In |
|          4 |  Tim | 19:10:00 | 01/06/2013_Out |
|          4 |  Tim | 11:49:00 |  01/20/2013_In |
|          4 |  Tim | 19:10:00 | 01/20/2013_Out |

获得此结果后，可以应用轴。如果您提前知道日期值，则可以对这些值进行硬编码，如下所示：

select *
from
(
  select employeeid, name, 
    convert(char(8), doortime, 108) DoorTime,
    date + '_'+ col as col_names
  from
  (
    select p.employeeid,
      p.name,
      convert(char(10),d.doordate, 101) date,
      min(d.doordate) [In],
      max(d.doordate) [Out]
    from person p
    left join doorlog d
      on p.employeeid = d.employeeid
    group by p.employeeid, p.name, 
      convert(char(10),d.doordate, 101)
  ) src
  unpivot
  (
    doortime
    for col in ([In], [Out])
  ) unpiv
) p
pivot
(
  max(doortime)
  for col_names in ([01/06/2013_In], [01/06/2013_Out],
                    [01/08/2013_In], [01/08/2013_Out],
                    [01/10/2013_In], [01/10/2013_Out],
                    [01/15/2013_In], [01/15/2013_Out],
                    [01/20/2013_In], [01/20/2013_Out],
                    [01/23/2013_In], [01/23/2013_Out],
                    [01/31/2013_In], [01/31/2013_Out])
) piv

看

但是对于您的情况，您可能需要使用动态SQL来生成结果，因为您很可能希望在任何一个月内都能动态生成结果。此的动态SQL版本是：

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT ',' + QUOTENAME(date +'_'+Logname) 
                    from
                    (
                      select doordate,
                        convert(char(10),doordate, 101) date, 
                        LogName
                      from DoorLog
                      cross apply
                      (
                        select 'In' LogName
                        union all
                        select 'Out' 
                      ) l
                    ) s   
                    group by convert(char(10), doordate, 112), date, Logname
                    order by convert(char(10), doordate, 112)
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query 
    = 'select employeeid, name, '+@cols+'
       from
       (
         select employeeid, name, 
          convert(char(8), doortime, 108) DoorTime,
          date + ''_''+ col col_names
         from
         (
            select p.employeeid,
              p.name,
              convert(char(10),d.doordate, 101) date,
              min(d.doordate) [In],
              max(d.doordate) [Out]
            from person p
            left join doorlog d
              on p.employeeid = d.employeeid
            group by p.employeeid, p.name, 
              convert(char(10),d.doordate, 101)
         )src
         unpivot
         (
           doortime
           for col in ([In], [Out])
         ) unpiv
       ) p
       pivot
       (
          max(doortime)
          for col_names in('+@cols+')
       ) piv'

execute(@query)

看

两个查询的结果都是：

| EMPLOYEEID | NAME | 01/06/2013_IN | 01/06/2013_OUT | 01/08/2013_IN | 01/08/2013_OUT | 01/10/2013_IN | 01/10/2013_OUT | 01/15/2013_IN | 01/15/2013_OUT | 01/20/2013_IN | 01/20/2013_OUT | 01/23/2013_IN | 01/23/2013_OUT | 01/25/2013_IN | 01/25/2013_OUT | 01/31/2013_IN | 01/31/2013_OUT |
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
|         11 |  Jim |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      11:15:00 |       16:25:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      12:31:00 |       17:50:00 |
|         10 | John |        (null) |         (null) |        (null) |         (null) |      09:21:00 |       15:45:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      10:31:00 |       16:45:00 |        (null) |         (null) |
|          8 | Mary |        (null) |         (null) |      01:29:00 |       02:25:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      13:29:00 |       18:25:00 |        (null) |         (null) |        (null) |         (null) |
|          4 |  Tim |      10:17:00 |       19:10:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      11:49:00 |       19:10:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |

---

此数据转换是一个复杂的过程。在SQL Server 2005+中，有一个函数可以为您旋转数据。有几种方法可以让你得到你想要的结果。两个版本都将实现

UNPIVOT

和

PIVOT

功能

样本数据：

CREATE TABLE Person ([EmployeeId] int, [Name] varchar(4));

INSERT INTO Person  ([EmployeeId], [Name])
VALUES
    (11, 'Jim'),
    (10, 'John'),
    (8, 'Mary'),
    (4, 'Tim');

CREATE TABLE DoorLog([EmployeeId] int, [DoorDate] datetime);

INSERT INTO DoorLog ([EmployeeId], [DoorDate])
VALUES
    (11, '2013-01-31 12:31:00'),
    (11, '2013-01-31 16:50:00'),
    (11, '2013-01-31 17:50:00'),
    (10, '2013-01-25 10:31:00'),
    (10, '2013-01-25 16:45:00'),
    (8, '2013-01-23 13:29:00'),
    (8, '2013-01-23 18:25:00'),
    (4, '2013-01-20 11:49:00'),
    (4, '2013-01-20 19:10:00'),
    (11, '2013-01-15 11:15:00'),
    (11, '2013-01-15 16:25:00'),
    (10, '2013-01-10 09:21:00'),
    (10, '2013-01-10 15:45:00'),
    (8, '2013-01-08 01:29:00'),
    (8, '2013-01-08 02:25:00'),
    (4, '2013-01-06 10:17:00'),
    (4, '2013-01-06 19:10:00');

您的查询首先获取每个日期具有最小/最大值的员工列表：

select p.employeeid,
  p.name,
  convert(char(10),d.doordate, 101) date,
  min(d.doordate) [In],
  max(d.doordate) [Out]
from person p
left join doorlog d
  on p.employeeid = d.employeeid
group by p.employeeid, p.name, 
  convert(char(10),d.doordate, 101)

看

下一步是

UNPIVOT

，它将获取输入/输出时间的单独列，并将它们放入多行：

select employeeid, name, 
      convert(char(8), doortime, 108) DoorTime,
      date + '_'+ col as col_names
from
(
  select p.employeeid,
      p.name,
      convert(char(10),d.doordate, 101) date,
      min(d.doordate) [In],
      max(d.doordate) [Out]
  from person p
  left join doorlog d
      on p.employeeid = d.employeeid
  group by p.employeeid, p.name, 
      convert(char(10),d.doordate, 101)
) src
unpivot
(
  doortime
  for col in ([In], [Out])
) unpiv

看。结果如下所示：

| EMPLOYEEID | NAME | DOORTIME |      COL_NAMES |
-------------------------------------------------
|          4 |  Tim | 10:17:00 |  01/06/2013_In |
|          4 |  Tim | 19:10:00 | 01/06/2013_Out |
|          4 |  Tim | 11:49:00 |  01/20/2013_In |
|          4 |  Tim | 19:10:00 | 01/20/2013_Out |

获得此结果后，可以应用轴。如果您提前知道日期值，则可以对这些值进行硬编码，如下所示：

select *
from
(
  select employeeid, name, 
    convert(char(8), doortime, 108) DoorTime,
    date + '_'+ col as col_names
  from
  (
    select p.employeeid,
      p.name,
      convert(char(10),d.doordate, 101) date,
      min(d.doordate) [In],
      max(d.doordate) [Out]
    from person p
    left join doorlog d
      on p.employeeid = d.employeeid
    group by p.employeeid, p.name, 
      convert(char(10),d.doordate, 101)
  ) src
  unpivot
  (
    doortime
    for col in ([In], [Out])
  ) unpiv
) p
pivot
(
  max(doortime)
  for col_names in ([01/06/2013_In], [01/06/2013_Out],
                    [01/08/2013_In], [01/08/2013_Out],
                    [01/10/2013_In], [01/10/2013_Out],
                    [01/15/2013_In], [01/15/2013_Out],
                    [01/20/2013_In], [01/20/2013_Out],
                    [01/23/2013_In], [01/23/2013_Out],
                    [01/31/2013_In], [01/31/2013_Out])
) piv

看

但是对于您的情况，您可能需要使用动态SQL来生成结果，因为您很可能希望在任何一个月内都能动态生成结果。此的动态SQL版本是：

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT ',' + QUOTENAME(date +'_'+Logname) 
                    from
                    (
                      select doordate,
                        convert(char(10),doordate, 101) date, 
                        LogName
                      from DoorLog
                      cross apply
                      (
                        select 'In' LogName
                        union all
                        select 'Out' 
                      ) l
                    ) s   
                    group by convert(char(10), doordate, 112), date, Logname
                    order by convert(char(10), doordate, 112)
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query 
    = 'select employeeid, name, '+@cols+'
       from
       (
         select employeeid, name, 
          convert(char(8), doortime, 108) DoorTime,
          date + ''_''+ col col_names
         from
         (
            select p.employeeid,
              p.name,
              convert(char(10),d.doordate, 101) date,
              min(d.doordate) [In],
              max(d.doordate) [Out]
            from person p
            left join doorlog d
              on p.employeeid = d.employeeid
            group by p.employeeid, p.name, 
              convert(char(10),d.doordate, 101)
         )src
         unpivot
         (
           doortime
           for col in ([In], [Out])
         ) unpiv
       ) p
       pivot
       (
          max(doortime)
          for col_names in('+@cols+')
       ) piv'

execute(@query)

看

两个查询的结果都是：

| EMPLOYEEID | NAME | 01/06/2013_IN | 01/06/2013_OUT | 01/08/2013_IN | 01/08/2013_OUT | 01/10/2013_IN | 01/10/2013_OUT | 01/15/2013_IN | 01/15/2013_OUT | 01/20/2013_IN | 01/20/2013_OUT | 01/23/2013_IN | 01/23/2013_OUT | 01/25/2013_IN | 01/25/2013_OUT | 01/31/2013_IN | 01/31/2013_OUT |
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
|         11 |  Jim |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      11:15:00 |       16:25:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      12:31:00 |       17:50:00 |
|         10 | John |        (null) |         (null) |        (null) |         (null) |      09:21:00 |       15:45:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      10:31:00 |       16:45:00 |        (null) |         (null) |
|          8 | Mary |        (null) |         (null) |      01:29:00 |       02:25:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      13:29:00 |       18:25:00 |        (null) |         (null) |        (null) |         (null) |
|          4 |  Tim |      10:17:00 |       19:10:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |      11:49:00 |       19:10:00 |        (null) |         (null) |        (null) |         (null) |        (null) |         (null) |

---

你能发布你的表格结构、样本数据和期望的结果吗？你的结果必须有多少列？@BlueFoots:我已经用样本数据编辑了这个问题。@HamletHakobyan:除了员工id，结果每天需要两列..一列显示输入时间，另一列显示输出时间。。同样，每天都需要两列作为每月的基础…你能发布你的表格结构吗，样本数据，然后是期望的结果？你的结果必须有多少列？@BlueFoots:我已经用样本数据编辑了这个问题。@HamletHakobyan:除了员工id，结果每天需要两列..一列显示输入时间，另一列显示输出时间。。同样地，每天都需要两个栏目作为每月的基础…尊重：）真的是一个详细而清晰的解释。。让我试试这个，然后再恢复。。谢谢你的努力……：）谢谢sql查询运行得非常完美。。但是我在使用动态查询时遇到了一些问题。。。我不清楚如何使用动态sql生成结果…@bluefeet…我一直在尝试在代码中使用此过程。。但是遇到了一个新问题。。希望能在这里寻求帮助。。尊敬：）真是一个精心设计、解释得很清楚的。。让我试试这个，然后再恢复。。谢谢你的努力……：）谢谢sql查询运行得非常完美。。但是我在使用动态查询时遇到了一些问题。。。我不清楚如何使用动态sql生成结果…@bluefeet…我一直在尝试在代码中使用此过程。。但是遇到了一个新问题。。希望能在这里寻求帮助。。