SQL面试题:求时间差之和(有重复不计)
编程学习 2021-07-05 14:37www.dzhlxh.cn编程入门
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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：
+------+-------+------------+------------+ | 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。
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