我有一个名为Rentals
的SQL Server 2005表:
RentalID
Book
Date
我想使用查询为每本书返回一年中每月租金的数量。
结果应如下所示:
+--------------------------------+-----+-----+-----+
| 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
答案 0 :(得分:3)
通过在聚合函数内部使用CASE表达式,可以更简单地编写。这个过程叫做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。您可以使用条件聚合来获取每月和每年的每本书的计数,而不是每列多次查询表。
答案 1 :(得分:0)
如果使用数据透视表,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。