Question

我有一张像这样的预订表：

booking_id,
date,
client,
sponsor

我正在尝试每月摘要：

SELECT 
  MONTH(date) AS M,
  Sponsor,
  Client,
  COUNT(booking_id) AS c
FROM booking
GROUP BY
 M, Sponsor, Client

现在我想知道客户在哪些日期预订。我尝试使用STUFF（）（在这篇文章中引用：Simulating group_concat MySQL function in Microsoft SQL Server 2005?）但它与group-by语句冲突。

根据请求提供样本数据。目前我有以下内容：

M       Sponsor     Client  c     
March   AB          y       3
March   FE          x       4
April   AB          x       2

期望的输出：

M       Sponsor     Client  c   dates
March   AB          y       3   12, 15, 18
March   FE          x       4   16, 19, 20, 21
April   AB          x       2   4, 8

数字是天数（例如12月3日，3月15日，3月18日）。在mysql中我会使用group_concat（date）来获取最后一列。

答案的大赞誉： - ）

Answer 1

SELECT [Month] = DATENAME(MONTH, M), Sponsor, Client, c,
  [dates] = STUFF((SELECT ', ' + RTRIM(DATEPART(DAY, [date])) 
    FROM dbo.booking AS b
    WHERE b.Sponsor = x.Sponsor
      AND b.Client = x.Client
      AND b.[date] >= x.M AND b.[date] < DATEADD(MONTH, 1, x.M) 
    ORDER BY [date]
    FOR XML PATH('')), 1, 2, '')
FROM 
(
  SELECT 
      M = DATEADD(MONTH, DATEDIFF(MONTH, '19000101', [date]), '19000101'),
      Sponsor,
      Client,
      COUNT(booking_id) AS c
    FROM dbo.booking
    GROUP BY DATEADD(MONTH, DATEDIFF(MONTH, '19000101', [date]), '19000101'),
      Sponsor,
      Client
) AS x
ORDER BY M, Sponsor, Client;

请注意，如果赞助商/客户的组合在同一天有两次预订，则该日期号码将在列表中出现两次。

编辑以下是我测试的方式：

DECLARE @booking TABLE
( 
  booking_id INT IDENTITY(1,1) PRIMARY KEY,
  [date] DATE,
  Sponsor VARCHAR(32),
  Client VARCHAR(32)
);

INSERT @booking([date], Sponsor, Client) VALUES
('20120312','AB','y'), ('20120315','AB','y'), ('20120318','AB','y'),
('20120316','FE','x'), ('20120319','FE','x'), ('20120321','FE','x'), ('20120320','FE','x'),
('20120404','AB','x'), ('20120408','AB','x');

SELECT [Month] = DATENAME(MONTH, M), Sponsor, Client, c,
  [dates] = STUFF((SELECT ', ' + RTRIM(DATEPART(DAY, [date])) 
    FROM @booking AS b
    WHERE b.Sponsor = x.Sponsor
      AND b.Client = x.Client
      AND b.[date] >= x.M AND b.[date] < DATEADD(MONTH, 1, x.M) 
    ORDER BY [date]
    FOR XML PATH('')), 1, 2, '')
FROM 
(
  SELECT 
      M = DATEADD(MONTH, DATEDIFF(MONTH, '19000101', [date]), '19000101'),
      Sponsor,
      Client,
      COUNT(booking_id) AS c
    FROM @booking
    GROUP BY DATEADD(MONTH, DATEDIFF(MONTH, '19000101', [date]), '19000101'), 
      Sponsor, 
      Client
) AS x
ORDER BY M, Sponsor, Client;

结果：

Month   Sponsor Client  c       dates
------- ------- ------- ------- --------------
March   AB      y       3       12, 15, 18
March   FE      x       4       16, 19, 20, 21
April   AB      x       2       4, 8

模仿group_concat（）与GROUP BY结合使用

1 个答案: