<div class="codebody" id="code82427"> --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 (PADINDEX = OFF,STATISTICSnorECOmpuTE = 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
