面试题四：手写sql

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;

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：

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;

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
    -- 1991111
    -- 1991212
    -- 1991313
    -- 1991414
    -- 1992121
    -- 1992222
    -- 1992323
    -- 1992424
    -- 查询结果
    -- 年一季度二季度三季度四季度
    -- 199111121314
    -- 199221222324select 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;