SQL Server 2005 GROUP BY 和 COUNT 查询每个月答案

【问题标题】：SQL Server 2005 GROUP BY and COUNT query for each monthSQL Server 2005 GROUP BY 和 COUNT 查询每个月
【发布时间】：2015-01-22 03:39:40
【问题描述】：

我有一个名为 Rentals 的 SQL Server 2005 表：

RentalID
Book
Date

我想使用查询返回每本书在给定年份每个月的租金。

结果应该是这样的：

+--------------------------------+-----+-----+-----+  
|              Book              | Jan | Feb | Mar |
+--------------------------------+-----+-----+-----+  
| Isaac Asimov - Foundation      |   2 |   5 |   3 |  
| H.G. Wells - War of the Worlds |   4 |   3 |   1 |  
| Frank Herbert - Dune           |   7 |   4 |   6 |
+--------------------------------+-----+-----+-----+

到目前为止我的查询：

SELECT
Book, 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=1 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=2 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=3 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=4 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=5 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=6 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=7 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=8 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=9 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=10 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=11 AND year(Date)=2011), 
(SELECT COUNT(*) FROM Rentals WHERE month(Date)=12 AND year(Date)=2011) 
FROM Rentals 
GROUP BY Book

【问题讨论】：

标签： sql sql-server sql-server-2005 pivot

【解决方案1】：

如果你使用pivot代码更容易维护，

SELECT 
    BOOK,
    [1] as Jan ,
    [2] as Feb,
    [3] as Mar,
    [4] as Apr,
    [5] as May,
    [6] as Jun,
    [7] as Jul,
    [8] as Aug,
    [9] as Sep,
    [10] as Oct,
    [11] as Nov,
    [12] as Dec 
FROM
(
    SELECT 
       BOOK , 
       DATEPART(MONTH,[DATE]) AS PER 
    FROM 
        Rentals 
    WHERE 
        DATEPART(YEAR,[DATE]) = 2014
) AS P PIVOT 
    (
    COUNT(PER) FOR PER IN ([1],[2],[3],[4],[5],[6],[7],[8],[9],[10],[11],[12])
    ) AS DATA

简单。

【讨论】：

【解决方案2】：

这可以通过在聚合函数中使用 CASE 表达式来更简单地编写。这个过程称为 PIVOT：

select book,
  sum(case when month(Date) = 1 then 1 else 0 end) Jan,
  sum(case when month(Date) = 2 then 1 else 0 end) Feb,
  sum(case when month(Date) = 3 then 1 else 0 end) Mar,
  sum(case when month(Date) = 4 then 1 else 0 end) Apr,
  sum(case when month(Date) = 5 then 1 else 0 end) May,
  sum(case when month(Date) = 6 then 1 else 0 end) Jun,
  sum(case when month(Date) = 7 then 1 else 0 end) Jul,
  sum(case when month(Date) = 8 then 1 else 0 end) Aug,
  sum(case when month(Date) = 9 then 1 else 0 end) Sep,
  sum(case when month(Date) = 10 then 1 else 0 end) Oct,
  sum(case when month(Date) = 11 then 1 else 0 end) Nov,
  sum(case when month(Date) = 12 then 1 else 0 end) Dec
from Rentals
where year(date) = 2011
group by book;

见SQL Fiddle with Demo。您无需为每一列多次查询表，而是使用条件聚合来获取每本书在月份和年份中的计数。

【讨论】：