• 数据分析MySQL阶段测验简答题


    有四张表

    用户表:student  列名:stu_id,stu_name

    学生分数表:student_score  列名:stu_id,score

    班级表:class  列名:class_id,class_name

    班级用户表:class_student  列名:class_id,stu_id

    使用SQL语句实现:

    1、实现学生的分数>80,并且<90的stu_id,stu_name,score

    2、实现获取班级名称为'实验班'的班级的平均分

    3、获取班级人数>40的班级,输出班级id,班级名称,班级人数

    解:

    1、

    mysql> select student.stu_id,student.stu_name,student_score.score from student
        -> left join student_score
        -> on student.stu_id = student_score.stu_id
        -> where student_score.score between 80 and 90;
    

    解析:

    用到二表左连,生成临时表格后输出 stu_id,stu_name,score三列,其中where子句使用between and语法限定条件>80<90的值

    2、

    mysql> select class.class_name,avg(score)
        ->  from class inner join class_student
        ->  on class.class_id = class_student.class_id
        ->  join student_score
        ->  on class_student.stu_id = student_score.stu_id
        ->  where class.class_name = '实验班';
    

    解析:

    用到三表内连,生成临时表格后输出class_name,avg(score)两列,其中where子句限定class_name='实验班'

    3、

    mysql> select class.class_id,class.class_name,count(class.class_name)
        ->  from class inner join class_student
        ->  on class.class_id = class_student.class_id
        ->  group by class.class_name
        ->  having count(class.class_name) > 40;
    

    解析:

    用到二表内连,生成临时表格后输出class_id,class_name,class_name的数量,其中having语句实现条件人数>40。注意,生成临时表格后,需要使用group by语句才能实现表格内容的分组,否则不能正确输出having后跟的count函数条件

    mysql初学,欢迎评论指正拍砖,谢谢!

  • 相关阅读:
    POJ 2068 Nim(博弈论)
    POJ 2311 Cutting Game (Multi-Nim)
    CodeForces 144B Meeting
    ZUFEOJ 2147 07染色带谜题
    CodeForces 779E Bitwise Formula
    CodeForces 779D String Game
    CodeForces 779C Dishonest Sellers
    CodeForces 779B Weird Rounding
    CodeForces 779A Pupils Redistribution
    HRBUST 1313 火影忍者之~静音
  • 原文地址:https://www.cnblogs.com/dmdoge/p/5499705.html
Copyright © 2020-2023  润新知