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SQL語句練習實例之一――找出最近的兩次晉升日期與工資額

2024-08-31 00:45:32
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復制代碼 代碼如下:


--程序員們在編寫一個雇員報表,他們需要得到每個雇員當前及歷史工資狀態的信息,
--以便生成報表。報表需要顯示每個人的晉升日期和工資數目。
--如果將每條工資信息都放在結果集的一行中,并讓宿主程序去格式化它。
--應用程序的程序員都是一幫懶人,他們需要在每個雇員的一行上得到當前
--和歷史工資信息。這樣就可以寫一個非常簡單的循環語句。
---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

----插入數據
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 連接進行查詢(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、這個方法是對每個雇員中的行進行編號,然后取出兩個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之后版本可以使用這種方法 ,使用CTE的方式來實現
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、使用視圖,將問題分為兩種情況

---1.只有一次工資變動的雇員

---2.有兩次或多次工資變動的雇員
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

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