编辑2:我发现问题是由SELECT子句中的聚合引起的,例如ROUND& COUNT,如果我删除ROUND&在视图表中成功显示所有行的COUNT。我需要使用两者,我仍然无法找到解决方案,有谁知道为什么它在使用ROUND&时改变结果COUNT以及如何解决这个问题?
查询来自一个名为view_caregiver的视图(只读)表,这些是涉及的表:
的护理人员
的地址
的 caregiver_review
SELECT `c`.`id` AS `id`,`c`.`user_id` AS `user_id`,`c`.`address_id` AS `address_id`,`c`.`first_name` AS `first_name`,`c`.`last_name` AS `last_name`,`c`.`email` AS `email`,`c`.`phone` AS `phone`,`c`.`status_id` AS `status_id`,`c`.`summary` AS `summary`,`c`.`about` AS `about`,`c`.`business_name` AS `business_name`,`c`.`medical_experience` AS `medical_experience`,`c`.`traveling_distance` AS `traveling_distance`,`c`.`accepting_new_patients` AS `accepting_new_patients`,`a`.`address1` AS `address1`,`a`.`address2` AS `address2`,`a`.`city` AS `city`,`a`.`state_id` AS `address_state`,`a`.`latitude` AS `latitude`,`a`.`longitude` AS `longitude`, COUNT(DISTINCT `cr`.`id`) AS `rating_count`, ROUND((AVG((((`cr`.`rating_knowledge` + `cr`.`rating_personality`) + `cr`.`rating_office`) + `cr`.`rating_timeliness`)) / 2),0) AS `rating`
FROM caregiver AS c
LEFT JOIN address AS a ON (c.address_id = a.id)
LEFT JOIN caregiver_review AS cr ON (c.id = cr.practice_id)
这样可行,但它只显示 caregiver_review 包含 practice_id 匹配的行,我希望它显示看护人表中的所有行即使没有 caregiver_review 匹配。我知道这可以通过LEFT JOIN完成,但我不明白为什么它不起作用!
答案 0 :(得分:2)
我认为您遇到的问题是由于在使用聚合函数时缺少GROUP BY。如果您没有GROUP BY功能,则不会返回所有数据。
你有两种方法可以做到这一点。您可以在子查询和GROUP BY中聚合一列:
SELECT `c`.`id` AS `id`,`c`.`user_id` AS `user_id`,
`c`.`address_id` AS `address_id`,`c`.`first_name` AS `first_name`,
`c`.`last_name` AS `last_name`,`c`.`email` AS `email`,
`c`.`phone` AS `phone`,`c`.`status_id` AS `status_id`,
`c`.`summary` AS `summary`,`c`.`about` AS `about`,
`c`.`business_name` AS `business_name`,
`c`.`medical_experience` AS `medical_experience`,
`c`.`traveling_distance` AS `traveling_distance`,
`c`.`accepting_new_patients` AS `accepting_new_patients`,
`a`.`address1` AS `address1`,`a`.`address2` AS `address2`,
`a`.`city` AS `city`,`a`.`state_id` AS `address_state`,
`a`.`latitude` AS `latitude`,`a`.`longitude` AS `longitude`,
cr.`rating_count`,
cr.`rating`
FROM caregiver AS c
LEFT JOIN address AS a
ON (c.address_id = a.id)
LEFT JOIN
(
select cr.practice_id,
COUNT(DISTINCT `cr`.`id`) AS `rating_count`,
ROUND((AVG((((`cr`.`rating_knowledge` + `cr`.`rating_personality`) + `cr`.`rating_office`) + `cr`.`rating_timeliness`)) / 2),0) AS `rating`
from caregiver_review AS cr
GROUP BY cr.practice_id
) cr
ON (c.id = cr.practice_id)
或者您可以在原始查询上进行聚合,并将GROUP BY应用于SELECT列表中未聚合的列:
SELECT `c`.`id` AS `id`,`c`.`user_id` AS `user_id`,
`c`.`address_id` AS `address_id`,`c`.`first_name` AS `first_name`,
`c`.`last_name` AS `last_name`,`c`.`email` AS `email`,
`c`.`phone` AS `phone`,`c`.`status_id` AS `status_id`,
`c`.`summary` AS `summary`,`c`.`about` AS `about`,
`c`.`business_name` AS `business_name`,
`c`.`medical_experience` AS `medical_experience`,
`c`.`traveling_distance` AS `traveling_distance`,
`c`.`accepting_new_patients` AS `accepting_new_patients`,
`a`.`address1` AS `address1`,`a`.`address2` AS `address2`,
`a`.`city` AS `city`,`a`.`state_id` AS `address_state`,
`a`.`latitude` AS `latitude`,`a`.`longitude` AS `longitude`,
COUNT(DISTINCT `cr`.`id`) AS `rating_count`,
ROUND((AVG((((`cr`.`rating_knowledge` + `cr`.`rating_personality`) + `cr`.`rating_office`) + `cr`.`rating_timeliness`)) / 2),0) AS `rating`
FROM caregiver AS c
LEFT JOIN address AS a
ON (c.address_id = a.id)
LEFT JOIN caregiver_review AS cr
ON (c.id = cr.practice_id)
GROUP BY `c`.`id`, `c`.`user_id`, `c`.`address_id`, `c`.`first_name`
, `c`.`last_name`, `c`.`email`, `c`.`phone`, `c`.`status_id`
, `c`.`summary`, `c`.`about`, `c`.`business_name`, `c`.`medical_experience`
, `c`.`traveling_distance`, `c`.`accepting_new_patients`
, `a`.`address1`, `a`.`address2`, `a`.`city`, `a`.`state_id`
, `a`.`latitude`, `a`.`longitude`
答案 1 :(得分:0)
你试过......
select c.id, a.address_id, cr.practice_id
from caregiver as c
left join address as a on ( c.address_id = a.id )
left join caregiver_review as cr on ( c.id = cr.practice_id )