MySQL，对嵌套计数感到困惑

Question

这是表格:

+------+--------+------+--------+
| sID  | sName  | GPA  | sizeHS |
+------+--------+------+--------+
|  123 | Amy    |  3.9 |   1000 |
|  234 | Bob    |  3.6 |   1500 |
|  345 | Craig  |  3.5 |    500 |
|  456 | Doris  |  3.9 |   1000 |
|  567 | Edward |  2.9 |   2000 |
|  678 | Fay    |  3.8 |    200 |
|  789 | Gary   |  3.4 |    800 |
|  987 | Helen  |  3.7 |    800 |
|  876 | Irene  |  3.9 |    400 |
|  765 | Jay    |  2.9 |   1500 |
|  654 | Amy    |  3.9 |   1000 |
|  543 | Craig  |  3.4 |   2000 |
+------+--------+------+--------+

我搞不懂这个查询背后的逻辑是什么

select *
from Student S1
where (select count(*) from Student S2
   where S2.sID <> S1.sID and S2.GPA = S1.GPA) =
  (select count(*) from Student S2
   where S2.sID <> S1.sID and S2.sizeHS = S1.sizeHS);

这是返回的内容:

+------+--------+------+--------+
| sID  | sName  | GPA  | sizeHS |
+------+--------+------+--------+
|  345 | Craig  |  3.5 |    500 |
|  567 | Edward |  2.9 |   2000 |
|  678 | Fay    |  3.8 |    200 |
|  789 | Gary   |  3.4 |    800 |
|  765 | Jay    |  2.9 |   1500 |
|  543 | Craig  |  3.4 |   2000 |
+------+--------+------+--------+

count 是一个聚合命令，它如何能够等于另一个聚合并在它是 where 条件时返回一个表？

Answer 1

count(*) 查询作为相关的子查询运行，它们都返回一个标量值(整数)。 您的主查询本身没有任何聚合。

两个 count(*) 查询返回两个数字，它们在完全合法的 where 条件下相互比较。

查询的计算结果如下:

select *
from Student S1
where (<count of students with the same GPA as this student>) 
        =
      (<count of students with the same sizeHS as this student>);

然后，例如，如果学生的计数(表中的一条记录)返回为 5 和 6，则 where 该记录的条件将被评估为:

select *
from Student S1
where 5 
        =
      6;