mysql 查询实战-变量方式-解答

news2024/11/27 4:18:54

对mysql 查询实战-变量方式-题目，进行一个解答。（先看题，先做，再看解答）

1、查询表中⾄少连续三次的数字

1，处理思路

要计算连续出现的数字，加个前置变量，记录上一个的值，跟当前值做比较，如果相同，则进行累加

SELECT *, @count:=IF(@pre=number,@count+1,1) AS count_num,
 @pre:=number AS pre 
FROM numbers,(SELECT@count:=0, @pre:=NULL) v;

统计count_num大于3的数字就好了。

2，进行统计

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;

2、查询销售额较昨⽇上升的记录

1，处理思路

要比较昨日上升的，要加个变量，记录昨天的值。用今天的值，跟昨天的值，做一个比较，并做一个标记。

 -- 根据时间排序，现在的值(amount)，比昨天的值（@pre）大的 flag设置为为1，否则为0
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;

再过来flag=1的数据

2，进行过滤统计

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

3、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，处理思路

排名，也是要跟前一个值做比较，值一样，排名不变；更小，就加1

-- 与前面一行进行比较，值一样，排名不变；更小，就加1
select name,votes,ranking from(
select *,@ranking:=if(votes=@pre,@ranking,@ranking+1) as ranking,
@pre:=votes as pre
from (select @ranking:=0,@pre:=null
) as v,vote order by votes desc
) as tmp;

这个排名看起来好奇怪，不符合实际的。两个第二名，下一个就是第四名了。

需要再做一次调整，比如两个第一后，累加的值就变成2了。因此需要再声明一个变量，用于判断累加的次数。

2，实际情况

-- 先累加相同的值， 然后把值赋值给另外一个变量。 再进行二次判断，如果sumSame的值大于0，就用sumSame。
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;

3，进行提取

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;

这样就符合实际的情况

4、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，先统计访问量大于1000

select *,@count:=if(visit_sum>=1000,@count+1,0) as count
from visit_summary;

2，用倒序排列，统计数量大于3的

再声明一个变量，用于记录上一条flag值

--连续 用倒序排列， 再声明一个变量，用于记录上一条flag值。
 SELECT *, @pre:=@flag, @flag:=IF((count_num>=3) AND count_num>0,1,0) AS flag
FROM(
SELECT *, @count:=IF(visit_sum>=1000, @count+1,0) AS count_num 
FROM visit_summary
) AS tmp1 ORDER BY id DESC;

由结果可以看出，对于小于3的，判断前面一条的flag是否为1，即@pre的值为1

3，对于小于3的，判断前面一条的flag是否为1

-- 小于3的，判断前面一条的flag是否为1。 
SELECT *, @pre:=@flag,
@flag:=IF((count_num>=3 OR @pre=1) AND count_num>0,1,0) AS flag
FROM(
SELECT *,@count:=IF(visit_sum>=1000,@count+1,0) AS count_num
 FROM visit_summary
) AS tmp ORDER BY id DESC;

然后再过滤flag值为1的，是符合要求的。

4，进行过滤

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;