35.HQL查询示例

[01].联合查询

select s.id,s.name,c.grade from stu s join cos c on s.id=c.id where c.cos_name="en" order by c.grade desc limit 3;

[02].查询英语成绩的平均值

select avg(grade) from cos where cos_name="en";

[03].复杂查询

select s.id,s.name,c.grade,c.avg_score,if(c.grade>c.avg_score,1,0) from stu s join (select id,grade,avg(grade) over(partition by cos_name) avg_score from cos where cos_name="en") c on s.id=c.id;