SQL面试题:求时间差之和(有重复不计)

时间:2020-12-15 开心学大数据人气:0

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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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