場景:假設有一個表,記錄了學生全部科目的成績,那麼如今要取出每一個科目分數最高的2位同窗的考試成績。
表名爲student_grade
表中字段爲:
course_id,course_name, student_id, student_name, grademysql
select course_id,course_name, student_id, student_name, grade, row_number() over(partition by course_id order by grade desc) as rank from student_grade where rank <= 2
注:MySQL中不支持使用row_number()函數sql
select * from ( select course_id, course_name, student_id, student_name, grade, (select count(*) from student_grade as t2 where t1.course_id=t2.course_id and t1.grade<=t2.grade) as rank from student_grade as t1) as t3 where rank <=2 order by course_id, rank
在嵌套查詢中,利用兩個表的course_id字段相等,能夠獲得這個分組下每條記錄在分組中的排名(rank)函數
相應的要獲取分組中,前n項的和或者平均值之類的,只須要以查詢的結果爲一個新的表,在該表上查詢相應的聚合值。優化
select sum(grade) from ( select course_id, course_name, student_id, student_name, grade, (select count(*) from student_grade as t2 where t1.course_id=t2.course_id and t1.grade<=t2.grade) as rank from student_grade as t1) as t3 where rank <=2 group by course_id
select *, grade/course_grade_sum as percent from( select course_id, course_name, student_id, student_name, grade, (select sum(grade) from student_grade as t2 where t1.course_id=t2.course_id) as course_grade_sum from student_grade as t1) as b
查詢每一個人考試分數最高的科目的記錄code
select * from student_grade as t1 where grade = (select max(grade) from student_grade as t2 where t1.student_id=t2.student_id)
select t1.student_id, t1.student_name, t1.course_id, t1.course_name, t1.grade from student_grade as t1 left join student_grade as t2 on t1.student_id= t2.student_id group by t1.student_id,t1.student_name, t1.course_id, t1.course_name, t1.grade having t1.grade=max(t2.grade)
最後的having條件中max(t2.grade)基於的範圍是什麼?是join的條件麼?排序
參考資料:慕課網MySQL開發技巧(一)開發
mysql中,能夠利用rand()函數排序,而後取前n條結果get
select * from table1 order by rand() limit n