Question

这里我要返回所有案例的总和。我的要求是了解特定产品是否在过去10周内销售。对于每周，当它在一周内销售时返回1或在没有任何销售时返回0。因此，如果一个产品在过去10周内都有销售，那么无论销售数量多少，我都必须返回10。但使用下面的查询，我得到的销售数量。

选择查询：

SELECT PartNo 
      , sum(CASE WHEN DocDate BETWEEN "2017-01-26" AND "2017-02-01" THEN 1 ELSE 0 END
         + CASE WHEN DocDate BETWEEN "2017-02-02" AND "2017-02-08" THEN 1 ELSE 0 END) as Qty1 
  FROM (`tbl_sales2`) 
 GROUP 
    BY `PartNo` 
 ORDER 
    BY `Qty1` DESC;

结果：

+-------------+------+
| PartNo      | Qty1 |
+-------------+------+
| 09343M05068 |    6 |
| 09343M05069 |    3 |
| 09343M05070 |    2 |
+-------------+------+
3 rows in set (0.00 sec)

预期结果：

+-------------+------+
| PartNo      | Qty1 |
+-------------+------+
| 09343M05068 |    2 |
| 09343M05069 |    2 |
| 09343M05070 |    1 |
+-------------+------+

表格查询：

CREATE TABLE IF NOT EXISTS `tbl_sales2` (
`SaleId` int(11) NOT NULL AUTO_INCREMENT,
`DocDate` date NOT NULL,
`PartNo` varchar(20) NOT NULL,
 PRIMARY KEY (`SaleId`)
 ) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=1;

示例数据：

INSERT INTO `tbl_sales2` (`SaleId`, `DocDate`, `PartNo`) VALUES
('1', '2017-01-26', '09343M05068'),
('2', '2017-01-27', '09343M05068'),
('3', '2017-01-28', '09343M05068'),
('4', '2017-01-29', '09343M05068'),
('5', '2017-01-28', '09343M05069'),
('6', '2017-01-29', '09343M05069'),
('7', '2017-02-02', '09343M05068'),
('8', '2017-02-03', '09343M05068'),
('9', '2017-02-02', '09343M05069'),
('10', '2017-02-02', '09343M05070'),
('11', '2017-02-02', '09343M05070');

Answer 1

如果MAX的给定日期范围内至少有一行，请使用SUM代替PartNo获取1：

select PartNo,
    max(case when DocDate between '2017-01-26'
                    and '2017-02-01' then 1 else 0 end) 
    + max(case when DocDate between '2017-02-02'
                    and '2017-02-08' then 1 else 0 end) as Qty1
from `tbl_sales2`
group by `PartNo`
order by `Qty1` desc;

产地：

PartNo      Qty1
09343M05068 2
09343M05069 2
09343M05070 1

计算mysql中所有案例的总和

1 个答案: