MYSQL左連接不顯示所有行

Question

編輯：我清理代碼以下，仍然沒有工作，它仍然只表示在caregiver_review包含匹配行（c.id = cr.practice_id），即使即時通訊使用LEFT JOIN

編輯2：我發現問題是由SELECT子句中的聚合引起的，例如ROUND和COUNT，如果我刪除ROUND和COUNT，則所有行都在視圖表中成功顯示。 我需要使用兩者，我仍然無法找到解決方案，是否有人知道為什么它在使用ROUND＆COUNT時改變結果以及我如何能夠正確解決這個問題？

該查詢來自一個名為view_caregiver的視圖（只讀）表，這些是涉及的表：
護理人員
地址
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)

這有效，但它只顯示caregiver_review包含practice_id匹配的行，我希望它顯示來自caregiver表的所​​有行，即使沒有caregiver_review匹配。 我知道這可以通過LEFT JOIN完成，但我不明白為什么它不起作用！

Answer 1

我認為你遇到的問題是由於在使用聚合函數時缺少GROUP BY。 如果您沒有GROUP BY功能，則不會返回所有數據。

你有兩種方法可以做到這一點。 您可以在子查詢中聚合，在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，您可以在原始查詢上進行聚合，並將GROUP BY應用於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`

Answer 2

你有沒有嘗試過...

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 )