求1SQL查询语句（最好是mysql）
日期：2014-05-18 浏览次数：20451 次
求一SQL查询语句（最好是mysql）
怎么把这样一个表儿
year month amount
1991 1 1.1
1991 2 1.2
1991 3 1.3
1991 4 1.4
1992 1 2.1
1992 2 2.2
1992 3 2.3
1992 4 2.4
查成这样一个结果
year m1 m2 m3 m4
1991 1.1 1.2 1.3 1.4
1992 2.1 2.2 2.3 2.4
------解决方案--------------------
这不是行列转换吗，a表示表名
SELECT
YEAR,
SUM(IF(MONTH=1,amount,0)) AS m1,
SUM(IF(MONTH=2,amount,0)) AS m2,
SUM(IF(MONTH=3,amount,0)) AS m3,
SUM(IF(MONTH=4,amount,0)) AS m4
FROM a
GROUP BY YEAR
------解决方案--------------------
SELECT
YEAR,
SUM(CASE MONTH WHEN 1 THEN amount ELSE 0 END) AS m1,
SUM(CASE MONTH WHEN 2 THEN amount ELSE 0 END) AS m2,
SUM(CASE MONTH WHEN 3 THEN amount ELSE 0 END) AS m3,
SUM(CASE MONTH WHEN 4 THEN amount ELSE 0 END) AS m4
FROM a
GROUP BY YEAR
用 case when zhen 的语法也可以，结果一样，这个数据是死的 ，如果数据是动态的就不行了。。
就必须用变量了
------解决方案--------------------
select a.year,sum(if(a.month='1',amount,'0')) as 'm1',
sum(if(a.month='2',amount,0)) as 'm2',
sum(if(a.month='3',amount,0)) as 'm3',
sum(if(a.month='4',amount,0)) as 'm4'
from row_column_test a
group by a.year;
------解决方案--------------------
怎么把这样一个表儿
year month amount
1991 1 1.1
1991 2 1.2
1991 3 1.3
1991 4 1.4
1992 1 2.1
1992 2 2.2
1992 3 2.3
1992 4 2.4
查成这样一个结果
year m1 m2 m3 m4
1991 1.1 1.2 1.3 1.4
1992 2.1 2.2 2.3 2.4
------解决方案--------------------
这不是行列转换吗，a表示表名
SELECT
YEAR,
SUM(IF(MONTH=1,amount,0)) AS m1,
SUM(IF(MONTH=2,amount,0)) AS m2,
SUM(IF(MONTH=3,amount,0)) AS m3,
SUM(IF(MONTH=4,amount,0)) AS m4
FROM a
GROUP BY YEAR
------解决方案--------------------
SELECT
YEAR,
SUM(CASE MONTH WHEN 1 THEN amount ELSE 0 END) AS m1,
SUM(CASE MONTH WHEN 2 THEN amount ELSE 0 END) AS m2,
SUM(CASE MONTH WHEN 3 THEN amount ELSE 0 END) AS m3,
SUM(CASE MONTH WHEN 4 THEN amount ELSE 0 END) AS m4
FROM a
GROUP BY YEAR
用 case when zhen 的语法也可以，结果一样，这个数据是死的 ，如果数据是动态的就不行了。。
就必须用变量了
------解决方案--------------------
select a.year,sum(if(a.month='1',amount,'0')) as 'm1',
sum(if(a.month='2',amount,0)) as 'm2',
sum(if(a.month='3',amount,0)) as 'm3',
sum(if(a.month='4',amount,0)) as 'm4'
from row_column_test a
group by a.year;
------解决方案--------------------
- SQL code
CREATE TABLE t1
(
nian INT,
yue INT,
amount DECIMAL(18,1)
)
INSERT INTO t1
SELECT 1991, 1, 1.1 UNION ALL
SELECT 1991, 2, 1.2 UNION ALL
SELECT 1991, 3, 1.3 UNION ALL
SELECT 1991, 4, 1.4 UNION ALL
SELECT 1992, 1, 2.1 UNION ALL
SELECT 1992, 2, 2.2 UNION ALL
SELECT 1992, 3, 2.3 UNION ALL
SELECT 1992, 4, 2.4
SELECT * FROM t1
DECLARE @str VARCHAR(8000)
SET @str='select nian'
SELECT @str=@str+',max(case when yue='+LTRIM(yue)+' then amount else 0 end) as ['+LTRIM(yue)+']'
FROM (SELECT DISTINCT yue FROM t1) AS a1
SET @str=@str+' from t1 group by nian'
PRINT @str
EXEC (@str)
-----------------
nian 1 2 3 4
1991 1.1 1.2 1.3 1.4
1992 2.1 2.2 2.3 2.4
------解决方案--------------------
字段名调整了下
- SQL code
DECLARE @str VARCHAR(8000)
SET @str='select nian'
SELECT @str=@str+',max(case when yue='+LTRIM(yue)+' then amount else 0 end) as [m'+LTRIM(yue)+']'
FROM (SELECT DISTINCT yue FROM t1) AS a1
SET @str=@str+' from t1 group by nian'
PRINT @str
EXEC (@str)
---------------------
nian m1 m2 m3 m4
1991 1.1 1.2 1.3 1.4
1992 2.1 2.2 2.3 2.4
免责声明： 本文仅代表作者个人观点，与爱易网无关。其原创性以及文中陈述文字和内容未经本站证实，对本文以及其中全部或者部分内容、文字的真实性、完整性、及时性本站不作任何保证或承诺，请读者仅作参考，并请自行核实相关内容。
