我有3个人，帐户和订阅。帐户属于Person一对多，帐户和订阅是多对多的。我需要一个可以从三个表中选择数据的SQL .-
人员表
+----------------------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+----------------------------+--------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| username | varchar(128) | NO | | NULL | |
| password | varchar(128) | NO | | NULL | |
| account_id | int(11) | NO | | NULL | |
+----------------------------+---------------+------+---------------+----------------+
+--------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+--------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| yearly_renew | tinyint(1) | NO | | NULL | |
| date_joined | date | NO | | NULL | |
| last_renewed | datetime | YES | | NULL | |
| signup | tinyint(1) | NO | | NULL | |
| reason | varchar(100) | YES | | NULL | |
| yearly_total | int(11) | YES | | NULL | |
| total | int(11) | YES | | NULL | |
+--------------+--------------+------+-----+---------+----------------+
帐户订阅表
+-----------------+---------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-----------------+---------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| account_id | int(11) | NO | MUL | NULL | |
| subscription_id | int(11) | NO | MUL | NULL | |
+-----------------+---------+------+-----+---------+----------------+
订阅表
+-------------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------------+--------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| subscription_type | varchar(150) | YES | | NULL | |
| hold | tinyint(1) | NO | | NULL | |
| start_date | date | YES | | NULL | |
| end_date | date | YES | | NULL | |
| amount_paid | double | YES | | NULL | |
| date_paid | date | YES | | NULL | |
| transaction_id | int(11) | YES | | NULL | |
| free | tinyint(1) | NO | | NULL | |
+-------------------+--------------+------+-----+---------+----------------+
在单个查询中期望输出 -
输出
+-----------+------------+--------------+
| person.id | account.id | subscription.id |
+-----------+------------+--------------+
| 10 | 11 | 20 |
| 15 | 32 | 45 |
| 23 | 43 | null |
+--------+---------+-----------------+
答案 0 :(得分：2)
试试这个。它可以正常工作......
SELECT p.Id AS person.id
,p.account_id AS account.id
,acsub.subscription_id as subscription.id
FROM Person AS p
LEFT JOIN Account-Subscription AS acsub ON p.account_id=acsub.account_id
答案 1 :(得分：0)
SELECT p.Id AS person.id
,p.account_id AS account.id
,acsub.subscription_id as subscription.id
FROM Person AS p
INNER JOIN
Account-Subscription AS acsub ON p.account_id=acsub.account_id