LeetCode SQL [Hard] 1384. 按年度列出销售总额

建表语句与原题:

Create table If Not Exists Product (product_id int, product_name varchar(30)) Create table If Not Exists Sales (product_id varchar(30), period_start date, period_end date, average_daily_sales int) Truncate table Product insert into Product (product_id, product_name) values (1, LC Phone ) insert into Product (product_id, product_name) values (2, LC T-Shirt) insert into Product (product_id, product_name) values (3, LC Keychain) Truncate table Sales insert into Sales (product_id, period_start, period_end, average_daily_sales) values (1, 2019-01-25, 2019-02-28, 100) insert into Sales (product_id, period_start, period_end, average_daily_sales) values (2, 2018-12-01, 2020-01-01, 10) insert into Sales (product_id, period_start, period_end, average_daily_sales) values (3, 2019-12-01, 2020-01-31, 1)

 

 

 

 

 

 

 本题知识点(MySQL):

1.时间函数

时间函数用的不多,用到查就好,本题里需要时间函数实现一年里的第一天,最后一天,以及两个日期之间相差的天数。

一年里的第一天:makedate(yy,days) ,yy 用当前年份,days代表一年的第几天。

一年的最后一天:CONCAT(yy,‘-12-31‘) ,用字符串拼接实现。

两个日期之间的时间差 : datediff(day1,day2)

2.一行转多行

先看一个例子:

SELECT SUBSTRING_INDEX( SUBSTRING_INDEX(A,B,C,D,E, ,, h.help_topic_id + 1), ,,- 1 ) AS chars FROM mysql.help_topic AS h WHERE h.help_topic_id < ( LENGTH(A,B,C,D,E) - LENGTH( REPLACE (A,B,C,D,E, ,, ‘‘ ) ) + 1 );
上面的SQL查出来的结果如下


mysql.help_topic 是mysql中自带的一张表,h.help_topic_id是自增id,从0开始。

理解了这个例子后本题就可以下手了。

MySQL版本题解:
SELECT sales.`product_id`, product.`product_name`, help_topic_id + YEAR(sales.`period_start`) AS report_year, DATEDIFF( IF( help_topic_id + YEAR(sales.`period_start`) < YEAR(sales.`period_end`), CONCAT( help_topic_id + YEAR(sales.`period_start`), -12-31 ), sales.`period_end` ), IF( help_topic_id = 0, sales.`period_start`, MAKEDATE( YEAR(sales.`period_start`) + help_topic_id, 1 ) ) ) * sales.`average_daily_sales` + sales.`average_daily_sales` AS total_amount FROM mysql.help_topic AS h, sales, Product WHERE h.help_topic_id < ( YEAR(sales.`period_end`) - YEAR(sales.`period_start`) + 1 ) AND Product.`product_id` = sales.`product_id`;