【SQL解惑】谜题15：找出最近两次工资

解惑一：

1、创建表和插入数据

create table Salaries

(

emp_name char(10) not null,

sal_date date not null,

sal_amt decimal (8,2) not null,

primary key (emp_name,sal_date)

)

insert into Salaries

values ('Tom','1996-06-20',500.00),

('Tom','1996-08-20',700.00),

('Tom','1996-10-20',800.00),

('Tom','1996-12-20',900.00),

('Dick','1996-06-20',500.00),

('Harry','1996-07-20',500.00),

('Harry','1996-09-20',700.00)

go

2、创建视图

create view Salaries2(emp_name,sal_date,sal_amt)

as select s0.emp_name,s0.sal_date,MAX(s0.sal_amt)

from Salaries as s0,Salaries as s1

where s0.sal_date <= s1.sal_date

and s0.emp_name = s1.emp_name

group by s0.emp_name,s0.sal_date

having COUNT(*) <= 2

go

解惑二：

select s0.emp_name,s0.sal_date,s0.sal_amt,s1.sal_date,s1.sal_amt

from Salaries as s0,Salaries as s1

where s0.emp_name = s1.emp_name

and s0.sal_date = (select MAX(s2.sal_date)

from Salaries as s2

where s0.emp_name = s2.emp_name)

and s1.sal_date = (select MAX(s3.sal_date)

from Salaries as s3

where s0.emp_name = s3.emp_name

and s3.sal_date < s0.sal_date)

union all

select s4.emp_name,max(s4.sal_date),MAX(s4.sal_amt),null,null

from Salaries as s4

group by s4.emp_name

having COUNT(*) = 1

解惑三：

1、语句块一（获取表的每个人的最大日期）：

（1）from子句获取Salaries表数据；

（2）groupby子句将表数据按emp_name分组

（3）select子句选择emp_name和最大的sal_date

select w.emp_name,MAX(w.sal_date) as maxdate

from Salaries as w

group by w.emp_name

2、语句块二（获取表的每个人的最大日期和第二大日期）：

（1）from子句获取上面语句块一的数据；

（2）left outer join子句左外连接表Salaries，当表a的emp_name与表x的emp_name且表a的日期（即最大日期）小于表x的日期时就添加到表a中；

（3）groupby子句将数据按emp_name和maxdate进行分组；

（4）select语句选择表a的emp_name，表a的maxdate（即最大的日期），表x的最大日期（即除了最大日期之外的第二大日期）

select a.emp_name,a.maxdate,max(x.sal_date) as maxdate2

from a

left outer join Salaries as x

on a.emp_name = x.emp_name and a.maxdate > x.sal_date

group by a.emp_name,a.maxdate

3、语句块三：

（1）from子句获取上面语句块二的数据；

（2）left outer join子句左外连接表Salaries，当表y与表z的emp_name和sal_date相等时则左外连接。

（3）select子句获取名字、最近日期、最近日期的工资、最近第二次的日期、最近第二次日期的工资

select b.emp_name,b.maxdate,y.sal_amt,b.maxdate2,z.sal_amt

from b

left outer join Salaries as y

on b.emp_name = y.emp_name and b.maxdate = y.sal_date

left outer join Salaries as z

on b.emp_name = z.emp_name and b.maxdate2 = z.sal_date

4、完整语句

select b.emp_name,b.maxdate,y.sal_amt,b.maxdate2,z.sal_amt

from (select a.emp_name,a.maxdate,max(x.sal_date) as maxdate2

from (select w.emp_name,MAX(w.sal_date) as maxdate

from Salaries as w

group by w.emp_name) as a

left outer join Salaries as x

on a.emp_name = x.emp_name and a.maxdate > x.sal_date

group by a.emp_name,a.maxdate) as b

left outer join Salaries as y

on b.emp_name = y.emp_name and b.maxdate = y.sal_date

left outer join Salaries as z

on b.emp_name = z.emp_name and b.maxdate2 = z.sal_date

解惑四：

1、使用解惑一的视图，通过日期的比较，只有前一个日期小于或等于后一个日期中的统计次数为小于等于2的即为前两个数。

例如：

日期为2017.1.1，2017.1.2，2017.1.3，2017.1.4。

那么按照公式的要求，自联结后日期进行比较并分组，那么存在如下分组。

2017.1.1<=2017.1.1	2017.1.2<=2017.1.2	2017.1.3<=2017.1.3	2017.1.4<=2017.1.4
2017.1.1<=2017.1.2	2017.1.2<=2017.1.3	2017.1.3<=2017.1.4
2017.1.1<=2017.1.3	2017.1.2<=2017.1.4
2017.1.1<=2017.1.4

通过上面的比对可了解到，只要分组成员少于或等于2的，那么对应的就是最近的两天

create view Salaries2(emp_name,sal_date,sal_amt)

as select s0.emp_name,s0.sal_date,MAX(s0.sal_amt)

from Salaries as s0,Salaries as s1

where s0.sal_date <= s1.sal_date

and s0.emp_name = s1.emp_name

group by s0.emp_name,s0.sal_date

having COUNT(*) <= 2

2、选择数据

（1）选择其中的最大的数据。

select s1.emp_name,s1.sal_date,s1.sal_amt,s2.sal_date,s2.sal_amt

from Salaries2 as s1,Salaries2 as s2

where s1.emp_name = s2.emp_name

and s1.sal_date > s2.sal_date

（2）选择其中数据量为1次的数据，即只有一天的数据。

select emp_name,MAX(sal_date),MAX(sal_amt),null,null

from Salaries2

group by emp_name

having COUNT(*) = 1

（3）将两种数据合并

select s1.emp_name,s1.sal_date,s1.sal_amt,s2.sal_date,s2.sal_amt

from Salaries2 as s1,Salaries2 as s2

where s1.emp_name = s2.emp_name

and s1.sal_date > s2.sal_date

union all

select emp_name,MAX(sal_date),MAX(sal_amt),null,null

from Salaries2

group by emp_name

having COUNT(*) = 1

解惑五：

1、语句块一：

（1）from子句获取表Salaries为表1的数据；

（2）innerjoin表Salaries为表2的数据，根据两表的emp_name和表2的sal_date进行自联接；

（3）当表2的sal_date数据为获取表Salaries为表4的数据，且表4的emp_name和表1的emp_name相等的时候，表4的sal_date都小于表1的每一条sal_date的时候，当上一步的数据为空值时，则直接获取表2的sal_date即可。

以下为具体例子：

（1）左边为表1和表2内连接的结果，右边为表4的数据；

（2）第一个，当s1_sal_date的数据1996-6-20与表4的所有数据进行比较，都是小于或大于表4的数据，因此没有满足的数据可以返回，因此取本身这条数据的s2_sal_date；

（3）第二个，当s1_sal_date的数据1996-10-20与表4的所有数据进行比较，存在数据1996-10-20大于1996-6-20和1996-8-20，因此再这两个数据中取最大值，选择对应的s2_sal_date即可。

总结：以下的语句是可以查找出，每一行日期与之邻近的上一个日期

select (...)

from Salaries as s1

inner join Salaries as s2

on s2.emp_name = s1.emp_name

and s2.sal_date = coalesce((select MAX(s4.sal_date)

from Salaries as s4

where s4.emp_name = s1.emp_name

and s4.sal_date < s1.sal_date),s2.sal_date)

2、语句块二：

（1）以语句块一的数据为基础；

（2）where子句中的子查询筛选数据：当表1的数据与表3的数据中的emp_name相等时时，如果表3的每一个sal_date日期都大于表1的sal_date日期时，则选择数据；

（3）当子查询没有任何数据时，才返回对应的sal_date。

当表1的1996-6-20与表3的所有sal_date比较时，存在1996-8-20、1996-10-20、1996-12-20大于表1的数据，因此在exists子查询中会返回数据；

当表1的1996-12-20与表3的所有sal_date比较时，不存在任何数据大于表1的数据，因此在exists子查询中不会返回任何数据；

同时where子句中的是not exists(true)时，则不返回数据；同时where子句中的是not exists(false)时，则返回数据。

select s1.emp_name as s1_emp_name,

s1.sal_date as s1_sal_date,

s1.sal_amt as s1_sal_amt,

s2.emp_name as s2_emp_name,

s2.sal_date as s2_sal_date,

s2.sal_amt as s2_sal_amt

from Salaries as s1

inner join Salaries as s2

on (...)

where not exists(select *

from Salaries as s3

where s3.emp_name = s1.emp_name

and s3.sal_date > s1.sal_date)

3、语句块三

（1）select子句选择表1的emp_name、sal_date、sal_amt

（2）如果表1的数据与表2的数据不相同时，则直接选择表2的数据；

如果表1的数据与表2的数据相同时，则证明该行数据就只有1个日期而已，则直接为空即可。

select s1.emp_name,s1.sal_date as curr_date,s1.sal_amt as curr_amt,

case when s2.sal_date <> s1.sal_date then s2.sal_date end as prev_date,

case when s2.sal_date <> s1.sal_date then s2.sal_amt end as prev_amt

from Salaries as s1

inner join Salaries as s2

on s2.emp_name = s1.emp_name

and s2.sal_date = coalesce((select MAX(s4.sal_date)

from Salaries as s4

where s4.emp_name = s1.emp_name

and s4.sal_date < s1.sal_date),s2.sal_date)

where not exists(select *

from Salaries as s3

where s3.emp_name = s1.emp_name

and s3.sal_date > s1.sal_date)

解惑六：

1、创建视图

（1）from子句获取表Salaries的数据

（2）leftouterjoin表Salaries，当表1的emp_name与表2的emp_name相同时，且表1的日期大于表2的日期。相当于获取所有日期与所有比该日期小的集合，如果没有则为空。

create view SalaryHistory(emp_name,curr_date,curr_amt,prev_date,prev_amt)

as

select s0.emp_name,s0.sal_date as curr_date,

s0.sal_amt as curr_amt,

s1.sal_date as prev_date,

s1.sal_amt as prev_amt

from Salaries as s0

left outer join Salaries as s1 on s0.emp_name = s1.emp_name

and s0.sal_date > s1.sal_date

go

2、查询数据

（1）from子句获取数据

（2）where子句筛选表1的最大日期，表1代表所有日期的最大日期，表2代表所有日期的邻近日期的最大日期。筛选出来的日期即为表的最大日期和最大的邻近日期。

select s0.emp_name,

s0.curr_date,

s0.curr_amt,

s0.prev_date,

s0.prev_amt

from SalaryHistory as s0

where s0.curr_date = (select MAX(curr_date)

from SalaryHistory as s1

where s0.emp_name = s1.emp_name)

and s0.prev_date = (select MAX(prev_date)

from SalaryHistory as s2

where s0.emp_name = s2.emp_name

or s0.prev_date is null)

解惑七：

1、查询语句

（1）子查询中，是使用Rank函数将表Salaries按emp_name分组且按sal_date降序排列

（2）查询语句中只有leftouterjoin表插入第二名的日期且选择第一日期即可。

with SalaryRanks(emp_name,sal_date,sal_amt,pos)

as

(

select emp_name,sal_date,sal_amt,

RANK() over(PARTITION BY emp_name order by sal_date desc)

from Salaries

)

select C.emp_name,

C.sal_date as curr_date,C.sal_amt as curr_amt,

P.sal_date as prev_date,P.sal_amt as prev_amt

from SalaryRanks as c

left outer join SalaryRanks as P

on P.emp_name = c.emp_name

and P.pos = 2

where c.pos = 1

解惑八：

1、查询语句

（1）子查询将表Salaries按emp_name分组按sal_date降序排列并赋值

（2）where子句筛序序号小于3的数据

（3）select子句根据序号1或者2来确定最近的一次和最近一次的前一次

select s1.emp_name,

MAX(case when rn = 1 then sal_date else null end) as curr_date,

MAX(case when rn = 1 then sal_amt else null end) as curr_amt,

MAX(case when rn = 2 then sal_date else null end) as prev_date,

MAX(case when rn = 2 then sal_amt else null end) as prev_amt

from (select emp_name,sal_date,sal_amt,

RANK() over (PARTITION BY emp_name order by sal_date desc)

from Salaries) as s1(emp_name,sal_date,sal_amt,rn)

where rn < 3

group by s1.emp_name

解惑九：

该方法与解惑七中一样

with cte(emp_name,sal_date,sal_amt,rn)

as

(

select emp_name,sal_date,sal_amt,

ROW_NUMBER() over (PARTITION by emp_name order by sal_date desc) as rn

from Salaries

)

select o.emp_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.emp_name = i.emp_name and i.rn = 2

where o.rn = 1

解惑十：

·1、该语句在MSSQL2010、MSSQL2012的版本才能运行

select emp_name,curr_date,curr_amt,prev_date,prev_amt

from (select emp_name,

sal_date as curr_date,

sal_amt as curr_amt,

MIN(sal_amt) over (partition by emp_name

order by sal_date desc

rows between 1 following and following)

as prev_date,

min(sal_amt) over (partition by emp_name

order by sal_date desc

rows between 1 following and following)

as prev_amt,

row_number() over (partition by emp_name

order by sal_date desc)

as rn

from Salaries) as dt

where rn = 1