一，行转列

先建立测试数据

if OBJECT_ID(‘week_income‘) is not null
drop table week_income
go
create table week_income
(
    employee varchar(10),
    weekname varchar(10),
    income int
)
go
insert into week_income
select ‘張三‘,‘星期一‘,1000 union all
select ‘張三‘,‘星期二‘,2000 union all
select ‘張三‘,‘星期三‘,1500 union all
select ‘張三‘,‘星期四‘,2000 union all
select ‘張三‘,‘星期五‘,3000 union all
select ‘張三‘,‘星期六‘,4000 union all
select ‘張三‘,‘星期日‘,5000 union all
select ‘李四‘,‘星期一‘,1000 union all
select ‘李四‘,‘星期二‘,2000 union all
select ‘李四‘,‘星期三‘,1500 union all
select ‘李四‘,‘星期四‘,2000 union all
select ‘李四‘,‘星期五‘,3000 union all
select ‘李四‘,‘星期六‘,4000 union all
select ‘李四‘,‘星期日‘,8000 

select * from week_income

1.1 用传统的case when 语法实现行转列。

select employee as ‘社員‘
      ,sum(case weekname when ‘星期一‘ then income else 0 end) as ‘星期一‘
      ,sum(case weekname when ‘星期二‘ then income else 0 end) as ‘星期二‘
      ,sum(case weekname when ‘星期三‘ then income else 0 end) as ‘星期三‘
      ,sum(case weekname when ‘星期四‘ then income else 0 end) as ‘星期四‘
      ,sum(case weekname when ‘星期五‘ then income else 0 end) as ‘星期五‘
      ,sum(case weekname when ‘星期六‘ then income else 0 end) as ‘星期六‘
      ,sum(case weekname when ‘星期日‘ then income else 0 end) as ‘星期日‘
from week_income
group by employee

1.2 用sqlserver2005以后新增加的pivot语法实现行转列。

pivot的语法如下：

SELECT <non-pivoted column>,
    [first pivoted column] AS <column name>,
    [second pivoted column] AS <column name>,
    ...
    [last pivoted column] AS <column name>
FROM
    (<SELECT query that produces the data>)
   AS <alias for the source query>
PIVOT
(
    <aggregation function>(<column being aggregated>)
FOR
[<column that contains the values that will become column headers>]
    IN ( [first pivoted column], [second pivoted column],
    ... [last pivoted column])
) AS <alias for the pivot table>
<optional ORDER BY clause>;

select employee  as ‘社員‘,[星期一],[星期二],[星期三],[星期四],[星期五],[星期六],[星期日] --step 3
from (select employee ,weekname,income from week_income) t -- step 2
pivot
(
　　　　--step 1
    sum(income) for weekname in([星期一],[星期二],[星期三],[星期四],[星期五],[星期六],[星期日])
) b

pivot分为3个步骤：

1，进行行列转换

sum(income) for weekname in([星期一],[星期二],[星期三],[星期四],[星期五],[星期六],[星期日])

注意：聚合函数sum等是决定重复行数据的处理方法，比如有两个星期一，就把这两个值合并了。 

2，定义检索数据源(select employee ,weekname,income from week_income) t

注意别名t不能省略。这里要特别注意employee这一列并没有在步骤1中出现，sqlserver会默认按employee进行分组，这个功能挺棒的。

3，选择结果集中的列，全部选择可以用*。

employee  as ‘社員‘,[星期一],[星期二],[星期三],[星期四],[星期五],[星期六],[星期日]

注意：可以对列名取别名，也可以只检索其中的部分列，而不必全部检索出所有列。

转换结果

二，列转行

以上面的转换结果作为数据表，我们先把结果导入到一张tmp的表中

select * into tmp from (
select employee  as ‘社員‘,[星期一],[星期二],[星期三],[星期四],[星期五],[星期六],[星期日]
from (select employee ,weekname,income from week_income) t
pivot
(
    sum(income) for weekname in([星期一],[星期二],[星期三],[星期四],[星期五],[星期六],[星期日])
) b
) c

select * from tmp 

然后，我们将星期一到星期日这几列转换到行上。

2.1 用传统的union all方法。

select 社員 as employee,‘星期一‘ as weekname, 星期一 as income from tmp union all
select 社員 as employee,‘星期二‘ as weekname, 星期二 as income from tmp union all
select 社員 as employee,‘星期三‘ as weekname, 星期三 as income from tmp union all
select 社員 as employee,‘星期四‘ as weekname, 星期四 as income from tmp union all
select 社員 as employee,‘星期五‘ as weekname, 星期五 as income from tmp union all
select 社員 as employee,‘星期六‘ as weekname, 星期六 as income from tmp union all
select 社員 as employee,‘星期日‘ as weekname, 星期日 as income from tmp
order by 社員

2.2 用sqlserver2005以后新增的unpivot方法。

select 社員 as employee, weekname,income
from (select 社員,[星期一],[星期二],[星期三],[星期四],[星期五],[星期六],[星期日] from tmp) a
unpivot
(
    income for weekname in ([星期一],[星期二],[星期三],[星期四],[星期五],[星期六],[星期日] )
) b

unpivot是pivot的逆操作，income和weekname都是需要创建的新列，income指代数据的值，weekname指代列名的值。

转换后的结果和我们最初建立的表一模一样

三，动态实现行列旋转

上面的方式，都是代码写了几个固定值旋转，没有实现动态列，扩展性不强。

要实现动态，主要使用拼sql的方式实现，以行转列的传统写法做个例子，其他的可以举一反三。

declare @sql as varchar(4000)
set @sql = ‘select employee as ‘‘社員‘‘‘

select @sql = @sql + ‘,sum(case weekname when ‘‘‘ + weekname + ‘‘‘ then income else 0 end) as ‘‘‘ + weekname + ‘‘‘‘
from (
    select distinct weekname from week_income) a
set  @sql =@sql + ‘ from week_income    ‘
set  @sql =@sql + ‘ group by employee    ‘
print @sql
exec (@sql)

结果也是这样