Question

Auth2.dbo.Accounts： account_id，login_name，密码，其他列...

Auth.dbo.Account： account_id，帐户，密码，其他栏目......

我想在＆＃34; Auth.dbo.Account ＆＃34;中插入帐户（account_id，帐户，密码）。到＆＃34; Auth2.dbo.Accounts ＆＃34; （account_id，login_name，password）并给出我想要的其他列值。

我希望这可行，但是一旦我在SQL Management studio中写道，我在＆＃34;附近出现了错误语法错误。从Auth.dbo.account＆＃34;

中选择account_id

INSERT [dbo].[Accounts] ([account_id], [login_name], [password],
       [referral_id], [referral_code], [pcbang], [block],
       [withdraw_remain_time], [age], [auth_ok],
       [last_login_server_idx], [event_code], [server_list_mask], 
       [result], [ip], [game_code], [gamecode], [login_event],
       [email], [security_a_1], [security_a_2], [security_a_3],
       [security_a_4], [security_q_1], [security_q_2], [security_q_3],
       [security_q_4], [votepoints], [cash], [country]) 
VALUES (select account_id from auth.dbo.Account, N'Imad', 
               N'3cfbbd2ae3c3e416c6d00a5a12ee60e8', NULL, NULL, 
               0, 0, NULL, -1997, 1, NULL, NULL, NULL, NULL, 
               N'::1', NULL, NULL, NULL, N'imad.lekal@outlook.com',
               NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, 
               0, 189, NULL)

Answer 1

尝试使用简单的INSERT/SELECT。在这种情况下，account_id将来自auth.dbo.Account，所有其他值将是常量。

INSERT Auth2.[dbo].[Accounts] ([account_id], [login_name], [password], [referral_id], [referral_code], [pcbang], [block], [withdraw_remain_time], [age], [auth_ok], [last_login_server_idx], [event_code], [server_list_mask], [result], [ip], [game_code], [gamecode], [login_event], [email], [security_a_1], [security_a_2], [security_a_3], [security_a_4], [security_q_1], [security_q_2], [security_q_3], [security_q_4], [votepoints], [cash], [country]) 
select account_id , N'Imad', N'3cfbbd2ae3c3e416c6d00a5a12ee60e8', NULL, NULL, 0, 0, NULL, -1997, 1, NULL, NULL, NULL, NULL, N'::1', NULL, NULL, NULL, N'imad.lekal@outlook.com', NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, 0, 189, NULL from auth.dbo.Account;

Answer 2

删除VALUES (

当您INSERT使用SELECT时，您不应使用VALUES，因为这些值来自SELECT

同样在表名前加上数据库名称，假设Auth和Auth2是2个不同的数据库。

最后，您的FROM已关闭。

以下应该工作。

INSERT Auth2.dbo.Accounts 
(
[account_id]
, [login_name]
, [password]
, [referral_id]
, [referral_code]
, [pcbang]
, [block]
, [withdraw_remain_time]
, [age]
, [auth_ok]
, [last_login_server_idx]
, [event_code]
, [server_list_mask]
, [result]
, [ip]
, [game_code]
, [gamecode]
, [login_event]
, [email]
, [security_a_1]
, [security_a_2]
, [security_a_3]
, [security_a_4]
, [security_q_1]
, [security_q_2]
, [security_q_3]
, [security_q_4]
, [votepoints]
, [cash]
, [country]
) 
SELECT 
account_id 
, N'Imad'
, N'3cfbbd2ae3c3e416c6d00a5a12ee60e8'
, NULL
, NULL
, 0
, 0
, NULL
, -1997
, 1
, NULL
, NULL
, NULL
, NULL
, N'::1'
, NULL
, NULL
, NULL
, N'imad.lekal@outlook.com'
, NULL
, NULL
, NULL
, NULL
, NULL
, NULL
, NULL
, NULL
, 0
, 189
, NULL
FROM auth.dbo.Account

专业提示：正确格式化您的SQL，有助于调试:)。

Answer 3

嗨，我认为这将首先帮助您创建数据库链接

CREATE DATABASE LINK your_db_link_name
 CONNECT TO Schema_Name
 IDENTIFIED BY <PWD>
 USING 'Your Database Name';

接下来你需要进行查询

INSERT [Accounts] ([account_id], [login_name], [password], [referral_id]
      , [referral_code], [pcbang], [block], [withdraw_remain_time], [age], [
      auth_ok], [last_login_server_idx], [event_code], [server_list_mask], [
      result], [ip], [game_code], [gamecode], [login_event], [email], [
      security_a_1], [security_a_2], [security_a_3], [security_a_4], [
      security_q_1], [security_q_2], [security_q_3], [security_q_4], [
      votepoints], [cash], [country])
      VALUES (
      select ac.account_id
      from Your_DB_Link_Name@Account ac, N'Imad', N'3cfbbd2ae3c3e416c6d00a5a12ee60e8',
      NULL, NULL, 0, 0, NULL, -1997, 1, NULL, NULL, NULL, NULL, N'::1', NULL,
      NULL, NULL, N'imad.lekal@outlook.com', NULL, NULL, NULL, NULL, NULL,
      NULL, NULL, NULL, 0, 189, NULL)

我认为如果您有任何疑问，这将对您有所帮助

从一个表插入到另一个表（不同的数据库）

3 个答案: