SQL - WHERE Condition on SUM()

Question

Is it possible to do something like this:

SELECT 
  `e`.*, 
  `rt`.`review_id`, 
  (SUM(vt.percent) / COUNT(vt.percent)) AS rating 
FROM `catalog_product_ent         


        

Answer 1

You need to use HAVING clause when filtering by aggregate functions:

SELECT 
  # `e`.*, => is this needed
  `rt`.`review_id`, 
  (SUM(vt.percent) / COUNT(vt.percent)) AS rating 
  AVG(vt.percent) AS rating1 # same as above 
FROM `catalog_product_entity` AS `e` 
INNER JOIN `rating_option_vote` AS `vt`
  ON vt.review_id = e.review_id 
GROUP BY `vt`.`review_id`
HAVING AVG(vt.percent) > 0

Two other notes:

1) SUM(x)/COUNT(x) <=> AVG(x)

2) You are including e.* in select but not in group by. MySql lets you do it but other databases wont.

Answer 2

You can use 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 vt.review_id = e.review_id 
GROUP BY `vt`.`review_id` 
HAVING rating >= 0;

As your question is tagged MySQL this solution should work, as the documentation shows:

...In standard SQL, a query that includes a GROUP BY clause cannot refer to nonaggregated columns in the HAVING clause that are not named in the GROUP BY clause. A MySQL extension permits references to such columns to simplify calculations. ...

This extension is enabled by default on official MySQL compilation.

Reference: http://dev.mysql.com/doc/refman/5.0/en/group-by-extensions.html (on the first occurrence of the word HAVING on the page)

Answer 3

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

Note: Added where to put ORDER BY statement

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

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

Answer 4

Use HAVING for aggregate conditions (such as the one you have)

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'
HAVING (SUM(vt.percent) / COUNT(vt.percent)) > 0