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这篇文章主要介绍了 SQL 如何求时间差之和,具有一定借鉴价值,感兴趣的朋友可以参考下,希望大家阅读完这篇文章之后大有收获,下面让丸趣 TV 小编带着大家一起了解一下。
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
brandall_daysnike13oppo12vivo18
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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