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SQL面試題:求時(shí)間差之和(有重復(fù)不計(jì))

 更新時(shí)間：2019年11月20日 09:11:32   作者：開心學(xué)大數(shù)據(jù)  
這篇文章主要介紹了SQL面試題:求時(shí)間差之和(有重復(fù)不計(jì))，文中通過示例代碼介紹的非常詳細(xì)，對(duì)大家的學(xué)習(xí)或者工作具有一定的參考學(xué)習(xí)價(jià)值，需要的朋友們下面隨著小編來一起學(xué)習(xí)學(xué)習(xí)吧

面試某某公司BI崗位的時(shí)候，面試題中的一道sql題，咋看一下很簡單，寫的時(shí)候發(fā)現(xiàn)自己缺乏總結(jié)，沒有很快的寫出來。

題目如下：

求每個(gè)品牌的促銷天數(shù)

表sale為促銷營銷表，數(shù)據(jù)中存在日期重復(fù)的情況，例如id為1的end_date為20180905，id為2的start_date為20180903，即id為1和id為2的存在重復(fù)的銷售日期，求出每個(gè)品牌的促銷天數(shù)(重復(fù)不算)

表結(jié)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 |
+------+-------+------------+------------+

最終結(jié)果應(yīng)為

brand all_days
nike 13
oppo 12
vivo 18

建表語句

-- ----------------------------
-- 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：

利用自關(guān)聯(lián)下一條記錄的方法

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

運(yùn)行結(jié)果

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

該方法對(duì)本題中的表格有效，但對(duì)于有id不連續(xù)的品牌的記錄時(shí)不一定適用。

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

運(yùn)行結(jié)果

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

其中條件

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

可以換成

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

結(jié)果同樣正確

用分析函數(shù)同樣可行的，自己電腦暫時(shí)沒裝oracle，用的mysql寫的。

以上就是本文的全部內(nèi)容，希望對(duì)大家的學(xué)習(xí)有所幫助，也希望大家多多支持腳本之家。

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