SQL多個包含連接

Question

下面是模式描述。我想構建一個查詢，對於給定用戶將返回所有通過case_users直接共享的案例或通過case_groups表間接共享。這裏是我的嘗試，我在那裏拉組用戶屬於前期：

SELECT * FROM `cases` 
INNER JOIN `case_users` ON `cases`.`id` = `case_users`.`case_id` 
INNER JOIN `case_groups` ON `cases`.`id` = `case_groups`.`case_id` 
WHERE `case_users`.`user_id` = '<USER_ID>' 
OR `case_groups`.`group_id` IN (<USER_GROUP_LIST>) 

EXPLAIN返回如下：Impossible WHERE noticed after reading const table...

我怎樣才能得到它呢？理想情況下，我希望一次性檢索所有案例 - 無需拉動USER_GROUP_LIST - 用戶所屬的組。

mysql> describe users; 
+-------------+--------------+------+-----+---------+----------------+ 
| Field  | Type   | Null | Key | Default | Extra   | 
+-------------+--------------+------+-----+---------+----------------+ 
| id   | int(11)  | NO | PRI | NULL | auto_increment | 
+-------------+--------------+------+-----+---------+----------------+ 

mysql> describe cases; 
+-------------+--------------+------+-----+---------+----------------+ 
| Field  | Type   | Null | Key | Default | Extra   | 
+-------------+--------------+------+-----+---------+----------------+ 
| id   | int(11)  | NO | PRI | NULL | auto_increment | 
+-------------+--------------+------+-----+---------+----------------+ 

mysql> describe case_users; 
+-------------+---------+------+-----+---------+-------+ 
| Field  | Type | Null | Key | Default | Extra | 
+-------------+---------+------+-----+---------+-------+ 
| user_id  | int(11) | NO | PRI | NULL |  | 
| case_id  | int(11) | NO | PRI | NULL |  | 
+-------------+---------+------+-----+---------+-------+ 

mysql> describe case_groups; 
+-------------+---------+------+-----+---------+-------+ 
| Field  | Type | Null | Key | Default | Extra | 
+-------------+---------+------+-----+---------+-------+ 
| case_id  | int(11) | NO | PRI | NULL |  | 
| group_id | int(11) | NO | PRI | NULL |  | 
+-------------+---------+------+-----+---------+-------+ 

mysql> describe group_users; 
+-------------+---------+------+-----+---------+-------+ 
| Field  | Type | Null | Key | Default | Extra | 
+-------------+---------+------+-----+---------+-------+ 
| group_id | int(11) | NO | PRI | NULL |  | 
| user_id  | int(11) | NO | PRI | NULL |  | 
+-------------+---------+------+-----+---------+-------+ 

Answer 1

您的加入只會返回其ID在case_users和case_groups這兩種情況下.. 如果一方或另一方，那麼你就需要2個查詢，您可以UNION獲得在一個單一的所有結果結果集：

SELECT `cases`.* FROM `cases` 
INNER JOIN `case_users` ON `cases`.`id` = `case_users`.`case_id` 
WHERE `case_users`.`user_id` = '<USER_ID>' 
UNION 
SELECT `cases`.* FROM `cases` 
INNER JOIN `case_groups` ON `cases`.`id` = `case_groups`.`case_id` 
WHERE `case_groups`.`group_id` IN (SELECT `group_users`.`group_id` 
            FROM `group_users` 
            WHERE `group_users`.`user_id` = '<USER_ID>')