本文介绍了SQL - SUM() 上的 WHERE 条件的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

是否可以这样做:

SELECT
  `e`.*,
  `rt`.`review_id`,
  (SUM(vt.percent) / COUNT(vt.percent)) AS rating
FROM `catalog_product_entity` AS `e`
INNER JOIN `rating_option_vote` AS `vt`
  ON vt.review_id = e.review_id
WHERE (rating >= '0')
GROUP BY `vt`.`review_id`

我特别想在除法结果值上加上 where 条件

In particular I would like to put a where condition on the division result value

推荐答案

这可以通过 HAVING 子句来完成:

This can be accomplished with a HAVING clause:

SELECT e.*, rt.review_id, (SUM(vt.percent) / COUNT(vt.percent)) AS rating
FROM catalog_product_entity AS e
INNER JOIN rating_option_vote AS vt ON e.review_id = vt.review_id
GROUP BY vt.review_id
HAVING (SUM(vt.percent) / COUNT(vt.percent)) >= 0
ORDER BY (SUM(vt.percent) / COUNT(vt.percent)) ASC

注意:添加了 ORDER BY 语句的放置位置

Note: Added where to put ORDER BY statement

查询优化器也不应该多次计算平均值,所以这里不应该担心.

The query optimizer should also not calculate the Average multiple times either, so that should not be a concern here.

正如@jagra 的回答中提到的,您应该能够使用 AVG() 而不是 SUM()/COUNT()

As was mentioned in @jagra's answer, you should be able to use AVG() instead of SUM() / COUNT()

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09-02 18:49