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  • mysql练习2

    14、查询每门课程被选修的次数

      SELECT course_id,COUNT(sid) FROM score GROUP BY course_id

    15、查询之选修了一门课程的学生姓名和学号

      SELECT sname FROM student WHERE sid in (
      SELECT student_id FROM score GROUP BY student_id HAVING COUNT(sid)=1)

    16、查询所有学生考出的成绩并按从高到低排序(成绩去重)

      SELECT DISTINCT num FROM score ORDER BY num DESC

    17、查询平均成绩大于85的学生姓名和平均成绩

      SELECT sname,AVG(num) FROM score INNER JOIN student ON score.student_id=student.sid
      GROUP BY student_id HAVING AVG(num)>85

    18、查询生物成绩不及格的学生姓名和对应生物分数

      SELECT sname 姓名,num 生物成绩 FROM student INNER JOIN (
      SELECT student_id,num FROM score WHERE course_id=(SELECT cid FROM course WHERE cname="生物") AND num<60)AS s
      ON student.sid=s.student_id

    19、查询在所有选修了李平老师课程的学生中,这些课程(李平老师的课程,不是所有课程)平均成绩最高的学生姓名

      

      SELECT sname FROM student INNER JOIN (

      SELECT MAX(s.a),s.student_id FROM (
       SELECT student_id,AVG(num) a FROM score WHERE course_id in (
        SELECT cid FROM course WHERE course.teacher_id=(
         SELECT tid FROM teacher WHERE tname="李平老师"))
       GROUP BY student_id
       ORDER BY AVG(num) DESC)AS s

      )AS s2 ON student.sid=s2.student_id

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  • 原文地址:https://www.cnblogs.com/guomeina/p/7252176.html
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