SQL语句练习实例之一——找出最近的两次晋升日期与工资额

复制代码 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','2011-6-20',2000 
union 
select 'Harry','2010-6-20',2000 
union 
select 'Harry','2011-6-20',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_date,sal_amt,rn) 
as 
( 
select name,sal_date,salary,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.sal_date, 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