SQL Server 2008 数据透视表按周分组答案

【问题标题】：SQL Server 2008 Pivot Table grouping by weekSQL Server 2008 数据透视表按周分组
【发布时间】：2015-02-26 13:07:50
【问题描述】：

我有这些数据，我需要使用诸如 PIVOT 表之类的东西以不同方式呈现。

我一直在尝试，但总是出错。

CREATE TABLE Table1
    ([Client_Id] varchar(8), [Project_Id] varchar(4), [WeekStart] datetime, [Total] int, [Country_Name] varchar(9));

INSERT INTO Table1
    ([Client_Id], [Project_Id], [WeekStart], [Total], [Country_Name])
VALUES
    ('CLIENTID', 'TEST', '2014-10-27 00:00:00', 1, 'UK'),
    ('CLIENTID', 'TEST', '2014-11-10 00:00:00', 1, 'UK'),
    ('CLIENTID', 'TEST', '2014-12-29 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-01-26 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-01-26 00:00:00', 2, 'Uruguay'),
    ('CLIENTID', 'TEST', '2015-01-26 00:00:00', 2, 'USA'),
    ('CLIENTID', 'TEST', '2015-01-26 00:00:00', 2, 'Italy'),
    ('CLIENTID', 'TEST', '2015-02-02 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-02-09 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-02-09 00:00:00', 2, 'Ireland'),
    ('CLIENTID', 'TEST', '2015-02-16 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-02-16 00:00:00', 2, 'Argentina'),
    ('CLIENTID', 'TEST', '2015-02-16 00:00:00', 2, 'Italy'),
    ('CLIENTID', 'TEST', '2015-02-16 00:00:00', 2, 'Brazil'),
    ('CLIENTID', 'TEST', '2015-02-16 00:00:00', 2, 'Mexico'),
    ('CLIENTID', 'TEST', '2015-02-16 00:00:00', 2, 'Spain'),
    ('CLIENTID', 'TEST', '2015-02-16 00:00:00', 1, 'Albania'),
    ('CLIENTID', 'TEST', '2015-02-16 00:00:00', 1, 'Portugal'),
    ('CLIENTID', 'TEST', '2015-03-02 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-03-30 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-04-27 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-06-01 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-06-29 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-08-03 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-08-31 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-09-28 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-11-02 00:00:00', 2, 'UK'),
    ('CLIENTID', 'TEST', '2015-11-30 00:00:00', 2, 'UK');

SQLFiddle

我需要以这种方式呈现数据

Client_Id        Project_IdCountry_Na10/27/201411/10/201412/29/2014 1/26/2015  2/2/2015  2/9/2015 2/16/2015  3/2/2015 3/30/2015 4/27/2015  6/1/2015 6/29/2015  8/3/2015 8/31/2015 9/28/2015 11/2/201511/30/2015
CLIENTID          TEST      UK                 1         1         2         2         2         2         2         2         2         2         2         2         2         2         2         2         2
CLIENTID          TEST      Uruguay                                          2
CLIENTID          TEST      USA                                              2
CLIENTID          TEST      Italy                                            2                             2
CLIENTID          TEST      Ireland                                                              2
CLIENTID          TEST      Argentina                                                                      2
CLIENTID          TEST      Brazil                                                                         2
CLIENTID          TEST      Mexico                                                                         2
CLIENTID          TEST      Spain                                                                          2
CLIENTID          TEST      Albania                                                                        1
CLIENTID          TEST      Portugal                                                                       1

任何帮助将不胜感激！

谢谢！

【问题讨论】：

您遇到了什么错误？你试过什么？
Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself. Questions without a clear problem statement are not useful to other readers
T-SQL - Pivot by week的可能重复

标签： sql-server sql-server-2008 pivot-table

【解决方案1】：

试试这个代码

DECLARE @Table1 TABLE
  (
     [Client_Id]    VARCHAR(8),
     [Project_Id]   VARCHAR(4),
     [WeekStart]    DATETIME,
     [Total]        INT,
     [Country_Name] VARCHAR(9)
  );

INSERT INTO @Table1
            ([Client_Id],
             [Project_Id],
             [WeekStart],
             [Total],
             [Country_Name])
VALUES      ('CLIENTID',
             'TEST',
             '2014-10-27 00:00:00',
             1,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2014-11-10 00:00:00',
             1,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2014-12-29 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-01-26 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-01-26 00:00:00',
             2,
             'Uruguay'),
            ('CLIENTID',
             'TEST',
             '2015-01-26 00:00:00',
             2,
             'USA'),
            ('CLIENTID',
             'TEST',
             '2015-01-26 00:00:00',
             2,
             'Italy'),
            ('CLIENTID',
             'TEST',
             '2015-02-02 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-02-09 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-02-09 00:00:00',
             2,
             'Ireland'),
            ('CLIENTID',
             'TEST',
             '2015-02-16 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-02-16 00:00:00',
             2,
             'Argentina'),
            ('CLIENTID',
             'TEST',
             '2015-02-16 00:00:00',
             2,
             'Italy'),
            ('CLIENTID',
             'TEST',
             '2015-02-16 00:00:00',
             2,
             'Brazil'),
            ('CLIENTID',
             'TEST',
             '2015-02-16 00:00:00',
             2,
             'Mexico'),
            ('CLIENTID',
             'TEST',
             '2015-02-16 00:00:00',
             2,
             'Spain'),
            ('CLIENTID',
             'TEST',
             '2015-02-16 00:00:00',
             1,
             'Albania'),
            ('CLIENTID',
             'TEST',
             '2015-02-16 00:00:00',
             1,
             'Portugal'),
            ('CLIENTID',
             'TEST',
             '2015-03-02 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-03-30 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-04-27 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-06-01 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-06-29 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-08-03 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-08-31 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-09-28 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-11-02 00:00:00',
             2,
             'UK'),
            ('CLIENTID',
             'TEST',
             '2015-11-30 00:00:00',
             2,
             'UK');

SELECT *
FROM   @Table1
       PIVOT (sum(total)
             FOR weekstart IN ([2014-10-27],
                               [2014-11-10],
                               [2014-12-29],
                               [2015-01-26],
                               [2015-02-02],
                               [2015-02-09],
                               [2015-02-16],
                               [2015-03-02],
                               [2015-03-30],
                               [2015-04-27],
                               [2015-06-01],
                               [2015-06-29],
                               [2015-08-03],
                               [2015-08-31],
                               [2015-09-28],
                               [2015-11-02],
                               [2015-11-30]) )t

【讨论】：

您可以删除 create table 和 insert 脚本，因为它已经存在问题