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>3时type=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;