面试题四:手写sql

  1. 矫正数据,有以下2个表,建表语句如下所示

    -- 订单表
    create table t_order
    (
        id    int auto_increment
            primary key,
        name  varchar(255) null,
        total int          null
    );
    -- 插入数据
    insert into sql_test.t_order (id, name, total) values (1, '家电', 1300);
    insert into sql_test.t_order (id, name, total) values (2, '洗漱', 170);
    insert into sql_test.t_order (id, name, total) values (3, '餐饮', 200);
    
    -- 详情表
    create table t_detail
    (
        id       int auto_increment
            primary key,
        detail   varchar(255) null,
        cost     int          null,
        order_id int          null
    );
    -- 插入数据
    insert into sql_test.t_detail (id, detail, cost, order_id) values (1, '洗衣机', 500, 1);
    insert into sql_test.t_detail (id, detail, cost, order_id) values (2, '电视机', 800, 1);
    insert into sql_test.t_detail (id, detail, cost, order_id) values (3, '牙膏', 100, 2);
    insert into sql_test.t_detail (id, detail, cost, order_id) values (4, '洗衣液', 70, 2);
    insert into sql_test.t_detail (id, detail, cost, order_id) values (5, '白菜', 200, 3);
    

    由于故障导致t_order表中的total值出现异常,使用一个sql语句进行矫正;

    update t_order o,
            (select order_id as oid, sum(cost) as t from t_detail GROUP BY order_id) b
    set o.total = b.t
    where o.id = b.oid;
    
  2. 分类求和题,有表如下

    create table t_type
    (
        id   int auto_increment
            primary key,
        type int null,
        num  int null
    );
    -- 插入数据
    insert into sql_test.t_type (id, type, num) values (1, 1, 100);
    insert into sql_test.t_type (id, type, num) values (2, 1, 200);
    insert into sql_test.t_type (id, type, num) values (3, 2, 500);
    insert into sql_test.t_type (id, type, num) values (4, 2, 200);
    insert into sql_test.t_type (id, type, num) values (5, 3, 300);
    insert into sql_test.t_type (id, type, num) values (6, 3, 180);
    insert into sql_test.t_type (id, type, num) values (7, 4, 50);
    insert into sql_test.t_type (id, type, num) values (8, 5, 60);
    insert into sql_test.t_type (id, type, num) values (9, 6, 70);
    

    要求:当type>3type=8,并且分类求和,要达到的效果如下:

    type sum
    1 300
    2 700
    3 480
    8 180

    实现sql语句,需要使用到case when xxx then xxx else xxx end语句:

    select case when type > 3 then 8 else type end as t, sum(case when type > 3 then num else num end)
    from t_type
    group by t;
    
  3. 学生成绩相关

    -- 学生表
    create table student(
    id int unsigned primary key auto_increment,
    name char(10) not null
    );
    insert into student(name) values('张三'),('李四');
    -- 课程表
    create table course(
    id int unsigned primary key auto_increment,
    name char(20) not null
    );
    insert into course(name) values('语文'),('数学');
    -- 学生成绩表
    create table student_course(
    sid int unsigned,
    cid int unsigned,
    score int unsigned not null,
    foreign key (sid) references student(id),
    foreign key (cid) references course(id),
    primary key(sid, cid)
    );
    insert into student_course values(1,1,80),(1,2,90),(2,1,90),(2,2,70);
    
    1. 查询重名的学生,按照name,id升序

      select id,name from student  where name in  (select name c from student group by name HAVING count(name) > 1)  order by name,id;
      
      -- exits写法
      select t.id,t.name from student t where EXISTS (select s.name from student s where s.name = t.name GROUP BY name HAVING count(s.name) > 1 ) order by t.name,t.id;
      
    2. 在student_course表中查询平均分不及格的学生,列出学生id和平均分

      select sid, AVG(score) as a from student_course GROUP BY sid HAVING a < 60;
      
    3. 在student_course表中查询每门课成绩都不低于80的学生id

      select DISTINCT sid from student_course where sid not in (select sid from student_course where score < 80);
      
    4. 查询每个学生的总成绩,结果列出学生姓名和总成绩

      select s.name, sum(c.score) from student_course c, student s  where c.sid = s.id GROUP BY sid;
      -- 上述方法会过滤掉没有成绩的人,因此需要使用左连接
      select name,sum(score)
      from student left join student_course
      on student.id=student_course.sid
      group by sid;
      
    5. 总成绩最高的学生,结果列出学生id和总成绩

      select sid, sum(score) as ss from student_course GROUP BY sid order by ss desc limit 1;
      
    6. 在student_course表查询课程1成绩第2高的学生,如果第2高的不止一个则列出所有的学生

      select * from student_coursewhere cid=1 and score = (
        select score from student_course where cid = 1 group by score order by score desc limit 1,1
      );
      
    7. 在student_course表查询各科成绩最高的学生,结果列出学生id、课程id和对应的成绩

      select * from student_course as x where score>=
      (select max(score) from student_course as y where cid=x.cid);
      
    8. 在student_course表中查询每门课的前2名,结果按课程id升序,同一课程按成绩降序

      select * from student_course x where
      2>(select count(distinct(score)) from student_course y where y.cid=x.cid and y.score>x.score)
      order by cid,score desc;
      
    9. 一个叫team的表,里面只有一个字段name,一共有4条纪录,分别是a,b,c,d,对应四个球队,两两进行比赛,用一条sql语句显示所有可能的比赛组合

      select a.name, b.name
      from team a, team b
      where a.name < b.name
      
    10. 竖变横

      -- 年	季度	销售
      -- 1991	1	11
      -- 1991	2	12
      -- 1991	3	13
      -- 1991	4	14
      -- 1992	1	21
      -- 1992	2	22
      -- 1992	3	23
      -- 1992	4	24
      -- 查询结果
      -- 年	一季度	二季度	三季度	四季度
      -- 1991	11	12	13	14
      -- 1992	21	22	23	24
      
      select 年, 
      sum(case when 季度=1 then 销售量 else 0 end) as 一季度, 
      sum(case when 季度=2 then 销售量 else 0 end) as 二季度, 
      sum(case when 季度=3 then 销售量 else 0 end) as 三季度, 
      sum(case when 季度=4 then 销售量 else 0 end) as 四季度 
      from sales group by 年;
      
posted @ 2021-03-17 13:05  一步一年  阅读(301)  评论(0编辑  收藏  举报