create database arron
go
use arron
go
www.2cto.com
-- createTable init Data
create table students (
name varchar(25),
class varchar(25),
grade int
)
insert into students values ('张三','语文',20)
insert into students values ('张三','数学',90)
insert into students values ('张三','英语',50)
insert into students values ('李四','语文',81)
insert into students values ('李四','数学',60)
insert into students values ('李四','英语',90)
-- solution1
select * from students
pivot(
max(grade)
FOR [class] IN ([语文],[数学],[英语])
) AS pvt www.2cto.com
-- solution2 相当于自连接
select A.Name,A.grade as 语文,B.grade as 数学,C.grade as 英语
from students A,students B,students C
where A.Name=B.Name and B.Name=C.Name
and A.class='语文' and B.class='数学'
and C.class='英语'
-- solution3
select name,
max(case when s.class='语文' then s.grade end) as 语文,
max(case when s.class='数学' then s.grade end) as 数学,
max(case when s.class='英语' then s.grade end) as 英语
from students s group by name
--在有id 的情况下
create table students2 (
id int primary key identity(1,1),
name varchar(25),
class varchar(25),
grade int
) www.2cto.com
insert into students2 values ('张三','语文',20)
insert into students2 values ('张三','数学',90)
insert into students2 values ('张三','英语',50)
insert into students2 values ('李四','语文',81)
insert into students2 values ('李四','数学',60)
insert into students2 values ('李四','英语',90)
-- 原先的solution1(有问题)
select * from students2
pivot(
max(grade)
FOR [class] IN ([语文],[数学],[英语])
) AS pvt
-- 原先的solution2 (ok)
select A.Name,A.grade as 语文,B.grade as 数学,C.grade as 英语
from students A,students2 B,students2 C
where A.Name=B.Name and B.Name=C.Name
and A.class='语文' and B.class='数学'
and C.class='英语'
-- 原先的solution3 （ok）
select name,
max(case when s.class='语文' then s.grade end) as 语文,
max(case when s.class='数学' then s.grade end) as 数学,
max(case when s.class='英语' then s.grade end) as 英语
from students s group by name www.2cto.com
--unpivot 函数使用
create table test1(id int,name varchar(20), Q1 int, Q2 int, Q3 int, Q4 int)
insert into test1 values(1,'a',1000,2000,4000,5000)
insert into test1 values(2,'b',3000,3500,4200,5500)
--实现的sql
select * from test1
select id ,[name],[jidu],[xiaoshou] from test1
unpivot
(
xiaoshou for jidu in
([q1],[q2],[q3],[q4])
)
as f
--- 以下的sql 可以替换上面的sql
select id,[name],
jidu='Q1',
xiaoshou=(select Q1 from test1 where id=a.id)
from test1 as a
union www.2cto.com
select id,[name],
jidu='Q2',
xiaoshou=(select Q2 from test1 where id=a.id)
from test1 as a
union
select id,[name],
jidu='Q3',
xiaoshou=(select Q3 from test1 where id=a.id)
from test1 as a
union
select id,[name],
jidu='Q4',
xiaoshou=(select Q4 from test1 where id=a.id)
from test1 as a
作者 maomao092092