mysql 查询实战2-解答

news2024/12/27 12:55:19

看了mysql 查询实战2-题目-CSDN博客的题目，继续进行解答。

6、查询⽹站访问⾼峰期

目标：查询网站访问高峰时期，高峰时期定义：至少连续三天访问量>=1000

1，关联查询

要连续三天，至少要声明“自身”三个去判断

select distinct a.* from visit_summary as a,visit_summary as b,visit_summary as c
where a.visit_sum>=1000 and b.visit_sum>=1000 and c.visit_sum>=1000 
and
(
a.id=b.id-1 and b.id=c.id-1
or
b.id=a.id-1 and a.id=c.id-1
or
b.id=c.id-1 and c.id=a.id-1
) order by id ;

2，用变量的方式处理

select * from(
select *,@flag:=if((count>=3 or @flag=1) and count>0,1,0) as flag
from(
  select *,@count:=if(visit_sum>=1000,@count+1,0) as count 
  from visit_summary
) as tmp order by id desc
) as tmp where tmp.flag=1 order by tmp.id;

如果是连续4天，5天，6天呢？这情况，用变量就很有优势了。用sql写，不断关联自身n次的话，sql就很繁杂了。

7、查询表中⾄少连续三次的数字

1，关联查询：

要连续三次，还是要声明“自身”三个去判断

select distinct a.number 
from numbers as a, numbers as b,numbers as c 
where a.id=b.id-1 and b.id=c.id-1 and a.number=b.number 
and b.number=c.number;

2，用变量的方式

 select distinct number from(
select *,@count:=if(@pre=number,@count+1,1) as count,@pre:=number as pre from
numbers, (select @count:=0,@pre:=null) v
) as tmp where tmp.count >= 3;

8、交换座位

目标：实现座位的交换，如果总数是奇数，那最后⼀⾏就不要交换了

1，异或判断

异或（XOR）操作通常用于二进制数据，可以通过使用^操作符来实现。异或运算符常用于比较不同位，即只有一位不同的值，返回1，否则返回0

select 0^1, 1^1, 2^1, 3^1, 4^1, 5^1;

2，关联查询

select a.seat_id, coalesce(b.name,a.name) as name 
from seat_plan a left join seat_plan b
on b.seat_id=(a.seat_id-1)^1+1
order by a.seat_id;

3，用取余的方式去处理

select (
  case when mod(seat_id,2)!=0 and seat_id!=total then seat_id+1
  when mod(seat_id,2)!=0 and seat_id=total then seat_id
  else seat_id-1
  end
) as id, name 
from seat_plan,(select count(1) as total from seat_plan) as tmp
order by id;

这个比较难想到，写起来也麻烦。

9、查询销售额较昨⽇上升的记录

1，关联自身进行比较

select b.* from sale as a,sale as b 
where datediff(b.record_date, a.record_date)=1 and
b.ammount > a.ammount;

2，用变量方式

 select id,record_date,ammount from
(
select *,@flag:=if(ammount>@pre,1,0) as flag,@pre:=ammount as pre from (select
@flag:=0,@pre:=null) as v, sale order by record_date
) as tmp where flag=1;

10、查询投票结果的排名情况

即第一名、第二名是谁，或者理解为，按倒序排好后，加个序号（区别就是同值的，排名是一样的）。

1，关联比较

要加一列，统计投票数大于当前行的

select *,(select count(b.votes) from vote as b where b.votes>=a.votes) as
ranking from vote as a order by votes desc;

2，用变量等方式

SELECT id, NAME, votes, ranking  FROM (
SELECT *, @same:=IF(votes=@pre, @same+1,0) AS same,  
 @ranking:=IF(votes=@pre, @ranking, @ranking+1+IF(@sumSame > 0, @sumSame,0)) AS ranking,
 @pre:=votes AS pre, @sumSame:=@same AS sunSame
FROM (SELECT @ranking:=0,@same:=0,@pre:=NULL,@sumSame:=NULL
) AS v,vote ORDER BY votes DESC
) AS temp;

这边排名，用变量的方式就显得比较繁琐了，用sql就简单很多。