Question

表由School_id和Student_name组成。

School_id   Student_Name
1           Raj
1           Tan
1           Brian
2           Raj
2           Brian
3           Raj
3           Tan

所以输出应该是Raj，因为Raj已经在所有学校学习过（1,2,3）。

Answer 1

如果已知不同学校的总数，则可以使用以下查询：

SELECT Student_Name
FROM mytable
GROUP BY Student_Name
HAVING COUNT(DISTINCT School_id) = 3

否则，您可以使用子查询代替3，如下所示：

SELECT COUNT(DISTINCT School_id)
FROM mytable

Answer 2

使用count（distinct ...）获取子查询中的学校数量，将此数字连接到外部查询，在该查询中计算学生与之关联的学校数量，并在having子句中进行过滤。

SELECT Student_Name, COUNT(DISTINCT School_id), t1.school_count
FROM yourtable
INNER JOIN (SELECT COUNT(DISTINCT School_id) as school_count FROM yourtable) t1
GROUP BY Student_Name
HAVING COUNT(DISTINCT School_id) = t1.school_count

Answer 3

SELECT Student_Name
FROM Table 
WHERE 
    COUNT(School_id) = (
        SELECT COUNT(*) 
        FROM Table 
        GROUP_BY(School_id)
    )
GROUP BY (Student_Name)

Answer 4

编辑：添加说明

简单解决您的确切要求

通过检查所有上过所有学校的学生来检索学生姓名（列表中不同学校的数量必须等于学生参加的不同学校的数量）

SELECT Student_Name
FROM Table1
GROUP BY Student_Name
HAVING Count(School_ID) =
  (SELECT Count(School_ID) 
   FROM (SELECT DISTINCT School_ID FROM Table1) t1
  )

MySQL Query为所有学校学习的学生

4 个答案: