欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页  >  IT编程

SQL面试题:求时间差之和(有重复不计)

程序员文章站 2022-06-07 10:01:08
面试某某公司bi岗位的时候,面试题中的一道sql题,咋看一下很简单,写的时候发现自己缺乏总结,没有很快的写出来。 题目如下: 求每个品牌的促销天数 表sale为促销营...

面试某某公司bi岗位的时候,面试题中的一道sql题,咋看一下很简单,写的时候发现自己缺乏总结,没有很快的写出来。

题目如下:

求每个品牌的促销天数

表sale为促销营销表,数据中存在日期重复的情况,例如id为1的end_date为20180905,id为2的start_date为20180903,即id为1和id为2的存在重复的销售日期,求出每个品牌的促销天数(重复不算)

表结果如下:

+------+-------+------------+------------+
| id | brand | start_date | end_date |
+------+-------+------------+------------+
| 1 | nike | 2018-09-01 | 2018-09-05 |
| 2 | nike | 2018-09-03 | 2018-09-06 |
| 3 | nike | 2018-09-09 | 2018-09-15 |
| 4 | oppo | 2018-08-04 | 2018-08-05 |
| 5 | oppo | 2018-08-04 | 2018-08-15 |
| 6 | vivo | 2018-08-15 | 2018-08-21 |
| 7 | vivo | 2018-09-02 | 2018-09-12 |
+------+-------+------------+------------+

最终结果应为

brand all_days
nike 13
oppo 12
vivo 18

建表语句

-- ----------------------------
-- table structure for sale
-- ----------------------------
drop table if exists `sale`;
create table `sale` (
 `id` int(11) default null,
 `brand` varchar(255) default null,
 `start_date` date default null,
 `end_date` date default null
) engine=innodb default charset=utf8;

-- ----------------------------
-- records of sale
-- ----------------------------
insert into `sale` values (1, 'nike', '2018-09-01', '2018-09-05');
insert into `sale` values (2, 'nike', '2018-09-03', '2018-09-06');
insert into `sale` values (3, 'nike', '2018-09-09', '2018-09-15');
insert into `sale` values (4, 'oppo', '2018-08-04', '2018-08-05');
insert into `sale` values (5, 'oppo', '2018-08-04', '2018-08-15');
insert into `sale` values (6, 'vivo', '2018-08-15', '2018-08-21');
insert into `sale` values (7, 'vivo', '2018-09-02', '2018-09-12');

方式1:

利用自关联下一条记录的方法

select brand,sum(end_date-befor_date+1) all_days from 
 (
 select s.id ,
  s.brand ,
  s.start_date ,
  s.end_date , 
  if(s.start_date>=ifnull(t.end_date,s.start_date) ,s.start_date,date_add(t.end_date,interval 1 day) ) as befor_date
 from sale s left join (select id+1 as id ,brand,end_date from sale) t on s.id = t.id and s.brand = t.brand
 order by s.id
 )tmp
 group by brand

运行结果

+-------+---------+
| brand | all_day |
+-------+---------+
| nike |  13 |
| oppo |  12 |
| vivo |  18 |
+-------+---------+

该方法对本题中的表格有效,但对于有id不连续的品牌的记录时不一定适用。

方式2:

select a.brand,sum(
 case 
  when a.start_date=b.start_date and a.end_date=b.end_date
  and not exists(
  select *
  from sale c left join sale d on c.brand=d.brand 
   where d.brand=a.brand
   and c.start_date=a.start_date
   and c.id<>d.id 
   and (d.start_date between c.start_date and c.end_date and d.end_date>c.end_date
   or 
  c.start_date between d.start_date and d.end_date and c.end_date>d.end_date)
    ) 
   then (a.end_date-a.start_date+1) 
  when (a.id<>b.id and b.start_date between a.start_date and a.end_date and b.end_date>a.end_date ) then (b.end_date-a.start_date+1)
  else 0 end
  ) as all_days 
from sale a join sale b on a.brand=b.brand group by a.brand

运行结果

+-------+----------+
| brand | all_days |
+-------+----------+
| nike |  13 |
| oppo |  12 |
| vivo |  18 |
+-------+----------+

其中条件

d.start_date between c.start_date and c.end_date and d.end_date>c.end_date
   or 
c.start_date between d.start_date and d.end_date and c.end_date>d.end_date

可以换成

c.start_date < d.end_date and (c.end_date > d.start_date)

结果同样正确

用分析函数同样可行的,自己电脑暂时没装oracle,用的mysql写的。

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持。