我有一张学生表:

id | age
--------
0  | 25
1  | 25
2  | 23

我想查询所有的学生,和一个额外的列,计算有多少学生是相同的年龄:

id | age | count
----------------
0  | 25  | 2
1  | 25  | 2
2  | 23  | 1

最有效的方法是什么?我担心子查询会很慢,我想知道是否有更好的方法。是吗?


当前回答

我会这样做:

select
 A.id, A.age, B.count 
from 
 students A, 
 (select age, count(*) as count from students group by age) B
where A.age=B.age;

其他回答

这应该可以工作:

您可以获取所有row表并计算行。

Select *,count(*) over() from students;

我会这样做:

select
 A.id, A.age, B.count 
from 
 students A, 
 (select age, count(*) as count from students group by age) B
where A.age=B.age;
select s.id, s.age, c.count
from students s
inner join (
    select age, count(*) as count
    from students
    group by age
) c on s.age = c.age
order by id

这应该可以工作:

SELECT age, count(age) 
  FROM Students 
 GROUP by age

如果你也需要id,你可以包括上面的子查询,像这样:

SELECT S.id, S.age, C.cnt
  FROM Students  S
       INNER JOIN (SELECT age, count(age) as cnt
                     FROM Students 
                    GROUP BY age) C ON S.age = C.age

如果你使用的是Oracle,那么一个叫做“分析”的功能就可以发挥作用。它是这样的:

select id, age, count(*) over (partition by age) from students;

如果你不使用Oracle,那么你需要加入到计数:

select a.id, a.age, b.age_count
  from students a
  join (select age, count(*) as age_count
          from students
         group by age) b
    on a.age = b.age