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感兴趣的小伙伴，下面一起跟随编程之家 52php.cn的小编两巴掌来看看吧！
代码如下:
--程序员们在编写一个雇员报表，他们需要得到每个雇员当前及历史工资状态的信息，
--以便生成报表。报表需要显示每个人的晋升日期和工资数目。
--如果将每条工资信息都放在结果集的一行中，并让宿主程序去格式化它。
--应用程序的程序员都是一帮懒人，他们需要在每个雇员的一行上得到当前
--和历史工资信息。这样就可以写一个非常简单的循环语句。
---示例：
create table salaries
( name nvarchar(50) not null,sal_date date not null,salary money not null,)
go
ALTER TABLE [dbo].salaries ADD CONSTRAINT [PK_salaries] PRIMARY KEY CLUSTERED
(
name,sal_date asc
)WITH (PAD_INDEX = OFF,STATISTICS_NORECOMPUTE = OFF,SORT_IN_TEMPDB = OFF,IGNORE_DUP_KEY = OFF,ONLINE = OFF,ALLOW_ROW_LOCKS = ON,ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
GO
----插入数据
insert into salaries
select 'TOM','2010-1-20',2000
union
select 'TOM','2010-6-20',2300
union
select 'TOM','2010-12-20',3000
union
select 'TOM','2011-6-20',4000
union
select 'Dick',2000
union
select 'Harry',2000
go
----方法一、使用left join 连接进行查询（sql 2000及以上版本）
select b.name,b.maxdate,y.salary,b.maxdate2,z.salary
from(select a.name,a.maxdate,MAX(x.sal_date) as maxdate2
from(select w.name,MAX(w.sal_date) as maxdate
from salaries as w
group by w.name) as a
left outer join salaries as x on a.name=x.name and a.maxdate>x.sal_date
group by a.name,a.maxdate) as b
left outer join salaries as y
on b.name=y.name and b.maxdate=y.sal_date
left outer join salaries as z
on b.name=z.name and b.maxdate2=z.sal_date
go
----方法二、这个方法是对每个雇员中的行进行编号，然后取出两个雇用日期最近的日期，
---（sql 2005以上版本）
select s1.name,MAX(case when rn=1 then sal_date else null end) as curr_date,MAX(case when rn=1 then salary else null end) as curr_salary,MAX(case when rn=2 then sal_date else null end) as prev_date,MAX(case when rn=2 then salary else null end) as curr_salary
from (select name,sal_date,salary,RANK() over(partition by name order by sal_date desc) rn
from salaries
) s1 where rn<3 group by s1.name
go
---方法三、在sql server 2005之后版本可以使用这种方法 ，使用CTE的方式来实现
with cte(name,sal_amt,rn)
as
(
select name,ROW_NUMBER() over(PARTITION by name order by sal_date desc) as rn from salaries
)
select o.name,o.sal_date AS curr_date,o.sal_amt as curr_amt,i.sal_date as prev_date,i.sal_amt as prev_amt from cte as o
left outer join cte as i on o.name=i.name and i.rn=2 where o.rn=1
go
----方法四、使用视图，将问题分为两种情况
---1.只有一次工资变动的雇员
---2.有两次或多次工资变动的雇员
create view v_salaries
as
select a.name,a.sal_date,MAX(a.salary) as salary from salaries as a,salaries as b
where a.sal_date<=b.sal_date and a.name=b.name group by a.name,a.sal_date
having COUNT(*)<=2
go
select a.name,a.salary,b.sal_date,b.salary from v_salaries a,v_salaries b
where a.name=b.name and a.sal_date>b.sal_date
union all
select name,max(sal_date),max(salary),cast(null as date),cast(null as decimal(8,2))
from v_salaries
group by name
having count(*)=1
go
drop table salaries
go
drop view v_salaries
（编辑：李大同）
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