Question

Name    Wage    Year
John    25,000  2016
John    35,000  2016
John    50,000  2015
Paul    70,000  2015
Paul    72,000  2016
George  50,000  2015
George  50,000  2016

假设我有上述SQL表，我如何比较一年只有一个工资的个人与有多个工资的个人的平均工资？例如。 2016年，一个工资的人比一个人的工资收入多，反之亦然？

我试图写一个只出现一年一次的人与那些出现不止一次的人的查询，并比较两组的平均工资。

我想要的结果是：

single_wage_average     multiple_wage_average     Year
50,000                  61,000                    2016

Answer 1

您可以使用平均值获得所有这些并计算如下：

select *, AverageWage = avg(wage) over(partition by name, [year]), 
    [CountOfOneWage]=count(name) over(partition by name, [year]) from #yourwage
order by name

使用此查询，您可以获得所需的输出。

您的输入表：

create table #yourwage (name varchar(20), wage money, year int)

insert into #yourwage 
( Name   ,  Wage  ,  [Year]  ) values
 ('John',    25000 , 2016   )
,('John',    35000 , 2016   )
,('John',    50000 , 2015   )
,('Paul',    70000 , 2015   )
,('Paul',    72000 , 2016   )
,('George',  50000 , 2015   )
,('George',  50000 , 2016   )

Answer 2

我将从聚合开始：

select name, year, avg(wage) as avg_wage, count(*) as num
from t
group by name, year;

然后你可以再次聚合：

select year,
       avg(case when num = 1 then avg_wage end) as singleton_avg,
       avg(case when num > 1 then avg_wage end) as duplicate_avg
from (select name, year, avg(wage) as avg_wage, count(*) as num
      from t
      group by name, year
     ) ny
group by year;

如何比较数据库中的不同值和重复值？

2 个答案: