Mysql返回id存在的表名

Question

我有这样的设计

accounts(id, username, email, password, ..)
admin_accounts(account_id, ...)
user_accounts(account_id, ....)
premium_accounts(account_id, ....)

id是帐户中的主键
account_id是外来的（帐户表上的引用ID）和这三个表中的主键（admin，user，premium）

知道id如何才能找到这个用户只有一个查询的类型？还知道id只能存在于三个表之一（admin，user，premium）

Answer 1

如果您使用UNION，可以尝试应用此处提出的解决方案 - mysql return table name

Answer 2

使用case：

select
  a.id,
  case 
    when aa.account_id is not null then 'admin_accounts'
    when ua.account_id is not null then 'user_accounts'
    when pa.account_id is not null then 'premium_accounts'
  else
    'No detail found'
  end as found_in
from
  accounts a
  left join admin_accounts aa on aa.account_id = a.id
  left join user_accounts ua on ua.account_id = a.id
  left join premium_accounts pa on pa.account_id = a.id
/*where -- In case you want to filter.
  a.id = ....*/

使用union

select
  id,
  found_in
from
    (select account_id as id, 'admin_accounts' as found_in 
    from admin_accounts aa 
    union all
    select account_id, 'user_accounts'
    from user_accounts ua 
    union all 
    select account_id, 'premium_accounts'
    from premium_accounts pa) a
/*where -- In case you want to filter.
  a.account_id = ....*/

Answer 3

您可以使用联合查询，例如

(select id, 'admin' as user_type from account inner join admin_accounts) UNION 
(select id, 'user' as user_type from account inner join user_accounts) UNION 
(select id, 'premium' as user_type from account inner join premium_accounts);