SQL interview question: Find the sum of time differences (ignore duplicates)

When I was interviewing for a BI position at a certain company, there was a SQL question in the interview. It seemed very simple at first glance, but when I was writing it, I found that I lacked summary and could not write it out quickly.

The topics are as follows:

Find the number of promotion days for each brand

Table sale is a promotional marketing table. There are repeated dates in the data. For example, the end_date of id 1 is 20180905, and the start_date of id 2 is 20180903. That is, id 1 and id 2 have repeated sales dates. Find the number of promotion days for each brand (duplicates are not counted)

The table results are as follows:

+------+-------+------------+------------+
| 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 |
+------+-------+------------+------------+

The final result should be

brand	all_days
Nike	13
OPPO	12
vivo	18

Create table statement

-- ----------------------------
-- 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');

Method 1:

Using the method of self-association to the next record

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 before_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

Operation Results

+-------+---------+
| brand | all_day |
+-------+---------+
| nike | 13 |
| oppo | 12 |
| vivo | 18 |
+-------+---------+

This method is valid for the table in this question, but may not be applicable to records of brands with discontinuous ids.

Method 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

Operation Results

+-------+----------+
| brand | all_days |
+-------+----------+
| nike | 13 |
| oppo | 12 |
| vivo | 18 |
+-------+----------+

Among the conditions

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

Can be replaced with

c.start_date < d.end_date AND (c.end_date > d.start_date)

The result is also correct

It is also feasible to use analytical functions. I don’t have Oracle installed on my computer yet, so I wrote it in MySQL.

The above is the full content of this article. I hope it will be helpful for everyone’s study. I also hope that everyone will support 123WORDPRESS.COM.

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