SQL按日期选择上个月组的总和

Question

我在 MySQL 数据中有下表

CREATE TABLE `new_table` (
  `date1` varchar(20) DEFAULT NULL,
  `organisation` varchar(45) DEFAULT NULL,
  `proportion` varchar(45) DEFAULT NULL,
  `id` int NOT NULL AUTO_INCREMENT,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci;

INSERT INTO `new_table` (`date1`, `organisation`, `proportion`) VALUES ('7/31/2020', 'Clinical Services', '50');
INSERT INTO `new_table` (`date1`, `organisation`, `proportion`) VALUES ('7/31/2020', 'Data and analytics', '60');
INSERT INTO `new_table` (`date1`, `organisation`, `proportion`) VALUES ('8/31/2020', 'Clinical Services', '12');
INSERT INTO `new_table` (`date1`, `organisation`, `proportion`) VALUES ('8/31/2020', 'Data and analytics', '40');
INSERT INTO `new_table` (`date1`, `organisation`, `proportion`) VALUES ('9/30/2020', 'Clinical Services', '25');
INSERT INTO `new_table` (`date1`, `organisation`, `proportion`) VALUES ('9/30/2020', 'Data and analytics', '20');

我想通过date1分别得到date1和proportion组的总和，分别为opening和closing，条件如下

1. 如果存在上个月的数据，则上个月的SUM(proportion) 为当月的opening，当月的SUM(proportion) 为closing。
2. 当上个月的数据不存在时，SUM(proportion) 是当月的开盘。

结果表应该是

Date	opening	closing
7/31/2020	110	110
8/31/2020	110	52
9/20/2020	52	45

我正在使用以下查询

SELECT
    date1 as 'Date',
    CASE
        WHEN (
            SELECT
                SUM(proportion)
            FROM
                new_table
            WHERE
                MONTH(str_to_date(date1, '%m/%d/%Y')) = MONTH(str_to_date(date1, '%m/%d/%Y')) - 1
            GROUP BY date1
            LIMIT 1
        ) > 0
        THEN (
            SELECT
                SUM(proportion)
            FROM new_table
            WHERE
                MONTH(str_to_date(date1, '%m/%d/%Y')) = MONTH(str_to_date(date1, '%m/%d/%Y')) - 1
            GROUP BY date1
            LIMIT 1
        )
        ELSE (
            SELECT
                sum(proportion)
            FROM new_table
            WHERE
                MONTH(str_to_date(date1, '%m/%d/%Y')) = MONTH(str_to_date(date1, '%m/%d/%Y'))
            GROUP BY date1
            LIMIT 1
        )
    END as 'Opening',
    SUM(proportion) as 'Closing'
FROM new_table
GROUP BY date1

但它给出了以下输出，其中Opening 对于所有行都是相同的。

示例：https://paiza.io/projects/Jgbxf3Xxag8fKIb3OuasjA?language=mysql

Answer 1

SELECT str_to_date(date1, '%m/%d/%Y') date1, 
       COALESCE(LAG(SUM(proportion)) OVER (ORDER BY str_to_date(date1, '%m/%d/%Y')), 
                SUM(proportion)) opening,
       SUM(proportion) closing
FROM new_table
GROUP BY date1

fiddle

如果月份中的日期可能不同，请使用LAST_DAY(str_to_date(date1, '%m/%d/%Y'))。

PS。使用了两个源数据值：取自问题文本和链接中的小提琴。

Answer 2

我已经重现了这种情况。您可以使用 Lag()Over() 来实现您正在寻找的内容：

select date1,coalesce( lag(sumprop,1)over( order by date1) ,sumprop) Opening,sumprop Closing  from (
select date1,sum(proportion) sumprop from new_table
group by date1)t

db-fiddle 链接：

https://www.db-fiddle.com/#&togetherjs=PieI81lvNT

对于旧版本的 MySQL：

select date1,coalesce((select sum(proportion) from new_table NT WHERE NT.DATE1<N.DATE1),sum(proportion)) Opening,
              sum(proportion)  Closing from new_table N
group by date1