【问题标题】：Pivot table for datetime日期时间的数据透视表
【发布时间】：2015-01-03 20:26:26
【问题描述】：

我有下表用于旋转。

示例：

表：

 create table testing
 (
  column_date datetime
 )

插入记录：

insert into testing values('2014-11-07'),('2014-11-08'),
('2014-11-01'),('2014-11-02'),('2014-11-04');

预期结果：

column_date  01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
----------------------------------------------------------------------------------------------------------
2014-11-07   0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
2014-11-08   0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
2014-11-01   1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
2014-11-02   0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
2014-11-04   0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0

尝试：

select a.column_date,[01],[02],[03],[04],[05],[06],[07],[08],[09],[10],
               [11],[12],[13],[14],[15],[16],[17],[18],[19],[20],
               [21],[22],[23],[24],[25],[26],[27],[28],[29],[30],[31]
from
(
    select column_date from testing
) as a
pivot
(
    count(column_date)
    for column_date in([01],[02],[03],[04],[05],[06],[07],[08],[09],[10],
               [11],[12],[13],[14],[15],[16],[17],[18],[19],[20],
               [21],[22],[23],[24],[25],[26],[27],[28],[29],[30],[31])
 ) pvt;

错误详情：

 Msg 8114, Level 16, State 1, Line 11
 Error converting data type nvarchar to date.
 Msg 473, Level 16, State 1, Line 11
 The incorrect value "01" is supplied in the PIVOT operator.
 Msg 4104, Level 16, State 1, Line 1
 The multi-part identifier "a.column_date" could not be bound.

【问题讨论】：

标签： sql sql-server tsql sql-server-2008-r2 pivot

【解决方案1】：

试试这个。您需要使用Datepart 在生成数据的SELECT 查询 中获取月份的天数，并在Pivot 中使用它来获取结果..

SELECT column_date,
       [01],[02],[03],[04],[05],[06],[07],[08],[09],
       [10],[11],[12],[13],[14],[15],[16],[17],[18],
       [19],[20],[21],[22],[23],[24],[25],[26],[27],
       [28],[29],[30],[31]
FROM   (SELECT column_date,
               Datepart(dd, column_date) dd,
               column_date               AS ddate
        FROM   #testing) AS a
       PIVOT ( Count(ddate)
             FOR dd IN( [01],[02],[03],[04],[05],[06],[07],[08],[09],
       [10],[11],[12],[13],[14],[15],[16],[17],[18],
       [19],[20],[21],[22],[23],[24],[25],[26],[27],
       [28],[29],[30],[31]) ) pvt;

输出

column_date             01  02  03  04  05  06  07  08  09  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31
---------------------------------------------------------------------------------------------------------------------------------------------------
2014-11-01 00:00:00.000 1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
2014-11-02 00:00:00.000 0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
2014-11-04 00:00:00.000 0   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
2014-11-07 00:00:00.000 0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
2014-11-08 00:00:00.000 0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0

【讨论】：