MySQL从另一个表中选择时自动增加插入到表中

时间:2013-07-21 15:18:52

标签: mysql select insert auto-increment

我有一个带有自动增量主键的表:

create table rt_table
(
  rtID int PRIMARY KEY AUTO_INCREMENT, 
  rt_user_id BIGINT,               /*user being retweeted*/
  rt_user_name varchar(70),        /*user name of rt_user_id*/
  source_user_id BIGINT,           /*user tweeting rt_user_id*/
  source_user_name varchar(70),    /*user name of source_user_id*/
  tweet_id BIGINT,                 /*fk to table tweets*/

  FOREIGN KEY (tweet_id) references tweets(tweet_id)
);

我希望从另一个表的部分填充此表:

insert into rt_table 
select rt_user_id, (select user_name from users u where u.user_id = t.rt_user_id),
       source_user_id, (select user_name from users u where u.user_id = t.source_user_id),
       tweet_id
  from tweets t
 where rt_user_id != -1;

我收到一个错误,指出列数不匹配,这是因为主键(这是一个自动递增的值,因此不需要设置)。我该如何解决这个问题?

2 个答案:

答案 0 :(得分:25)

您需要明确列出insert语句中的列:

insert into rt_table (rt_user_id, rt_user_name, source_user_id, source_user_name, tweet_id)
select rt_user_id, (select user_name from users u where u.user_id = t.rt_user_id),
       source_user_id, (select user_name from users u where u.user_id = t.source_user_id),
       tweet_id
  from tweets t
 where rt_user_id != -1;

另外,我认为使用显式连接是更好的形式,而不是嵌套选择:

insert into rt_table (rt_user_id, rt_user_name, source_user_id, source_user_name, tweet_id)
    select t.rt_user_id, u.user_name, t.source_user_id, su.user_name, t.tweet_id
    from tweets t left outer join
         users u
         on t.rt_user_id = u.user_id left outer join
         users su
         on t.source_user_id = su.user_id
    where rt_user_id != -1;

这通常(但并非总是)有助于优化器找到最佳查询计划。

答案 1 :(得分:20)

您只需在插入过程中将主键设置为NULL

INSERT INTO rt_table 
SELECT 
  NULL,
  rt_user_id,
  (SELECT 
    user_name 
  FROM
    users u 
  WHERE u.user_id = t.rt_user_id),
  source_user_id,
  (SELECT 
    user_name 
  FROM
    users u 
  WHERE u.user_id = t.source_user_id),
  tweet_id 
FROM
  tweets t 
WHERE rt_user_id != - 1 ;
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