SQL count命令，avg命令

with student as(
select 1 as id from dual union all
select null as id from dual union all
select 3 as id from dual 
)

--如果使用count(1)則計算所有，如果列出的使用count(列名)則會濾掉null值
select count(id), count(1),count(*) from student

 

 

with student as(
select 1 as id from dual union all
select null as id from dual union all
select 3 as id from dual 
)
select min(id), max(id) from student;

 

with student as(
select 1 as id from dual union all
select null as id from dual union all
select 3 as id from dual union all
select 3 as id from dual 
)
select count(distinct id) from student;

 

這裏在count中看到的是【去掉null 以後，非重複的內容】

 

 

-------------------------------------------------------------------------------------------------

with student as(
select 1 as id from dual union all
select null as id from dual union all
select 3 as id from dual union all
select 3 as id from dual 
)
select avg(id) from student;

 

結果是7/3的結果，即濾掉NULL以後，使用(1+3+3)/3作爲AVG的運算方式

但換個方法

with student as(
select 1 as id from dual union all
select null as id from dual union all
select 3 as id from dual union all
select 3 as id from dual 
)
select avg(nvl(id,0)) from student;

 

就是以4爲除數了。

 

 那麼，對於【組函數】我們可以使用下面這一個語句

 

with student as(
select 1 as id from dual union all
select null as id from dual union all
select 3 as id from dual union all
select 3 as id from dual 
)
select min(id),max(id),avg(nvl(id,0)),count(1) from student;