Question

我想对SQL中的两列求和，如果总和大于0，则将其输出到其中一列，如果小于0，则输出到另一列。

我的代码如下：

SELECT IF(SUM(`processed_quantity_long`,-1*`processed_quantity_short`) > 0,SUM(`processed_quantity_long`,-1*`processed_quantity_short`) AS `Position Long`,SUM(`processed_quantity_long`,-1*`processed_quantity_short`) AS `Position Short`) 
From table A 
GROUPBY date

它向我返回此错误：

您的SQL语法有错误；检查与您的MariaDB服务器版本相对应的手册以获取正确的语法，以便在第5行的'-1 * {processed_quantity_short）> 0，SUM（processed_quantity_long，-1 *`processed_qua'附近使用

不确定如何解决此错误。

Answer 1

SUM不需要两个参数。只需在计算总和之前减去两个数字即可：

SELECT
    CASE WHEN SUM(processed_quantity_long - processed_quantity_short) >= 0 THEN SUM(processed_quantity_long - processed_quantity_short) END AS `Position Long`,
    CASE WHEN SUM(processed_quantity_long - processed_quantity_short) <  0 THEN SUM(processed_quantity_long - processed_quantity_short) END AS `Position Short`
FROM tablea
GROUP BY date

Answer 2

如果

，您在里面做错了

 SELECT IF(
          SUM(processed_quantity_long-processed_quantity_short) > 0,
          SUM(`processed_quantity_long`-`processed_quantity_short`) ,
          SUM(`processed_quantity_long`-`processed_quantity_short`)  AS  `Position Short`
    ) From tableA GROUP BY date

常规if语句如下

SELECT IF(500<1000, "YES", "NO")

或用例

case when SUM(`processed_quantity_long`-`processed_quantity_short`) > 0
 then SUM(`processed_quantity_long`-`processed_quantity_short`)
 else SUM(`processed_quantity_long`-`processed_quantity_short`) as position
from tableA GROUP BY date

Answer 3

也许编写代码的最简单方法是：

SELECT GREATEST(SUM(processed_quantity_long - processed_quantity_short), 0) AS Position_Long,
       LEAST(SUM(processed_quantity_long - processed_quantity_short), 0) AS Position_Short
FROM tablea
GROUP BY date

对SQL中的两列求和，其中一列需要取反

3 个答案: