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:
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+ ------+-------+------------+------------+
| 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
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-- ----------------------------
-- 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
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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
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+ -------+---------+
| 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:
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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
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运行结果
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+ -------+----------+
| brand | all_days |
+ -------+----------+
| nike | 13 |
| oppo | 12 |
| vivo | 18 |
+ -------+----------+
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其中条件
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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
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可以换成
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c.start_date < d.end_date AND (c.end_date > d.start_date)
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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服务器之家。
原文链接:https://blog.csdn.net/u012955829/article/details/102754141