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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：
? 1 2 3 4 5 6 7 8 9 10 11+------+-------+------------+------------+
| 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
建表语句
? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21-- ----------------------------
-- 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
? 1 2 3 4 5 6 7 8 9 10 11select 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
? 1 2 3 4 5 6 7+-------+---------+
| 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：
? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18SELECT 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
? 1 2 3 4 5 6 7+-------+----------+
| brand | all_days |
+-------+----------+
| nike | 13 |
| oppo | 12 |
| vivo | 18 |
+-------+----------+
其中条件
? 1 2 3d.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
? 1c.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。
原文链接：https://blog.csdn.net/u012955829/article/details/102754141