mySQL - 比较表行

Question

我只是不太确定这样做的正确方法。我基本上有2张 table 。两个表具有相同的结构。 Database1(当前数据平均20k行)，Database2(数据累积，在任何给定时间可以超过100万行)。

我想比较两个表，并通过特定的时间戳查询给出 Database1 中不存在于 Database2 中的结果。我已经尝试过但查询需要很长时间才能搜索。

尝试过:

select distinct player from Database1
where not exists (select player from Database2 where snap = 1340981695)

SELECT Database1.player FROM Database1
INNER JOIN Database2 ON Database1.player NOT IN (Database2.player) AND Database2.snap = 1340981695
GROUP BY Database1.player

select distinct Database1.player from Database1
left join Database2 on Database1.player not in (Database2.player)
and Database2.snap = 1340981695

我仍然无法理解这个问题。感谢您的帮助。

Answer 1

共有三种方法。大致顺序为performance ，从最好到最差:

1. 使用外部联接:

   SELECT Database1.*
   FROM   Database1
     LEFT JOIN Database2
            ON Database1.id = Database2.id AND Database2.snap = 1340981695
   WHERE  Database2.id IS NULL
   

2. 使用IN:

   SELECT *
   FROM   Database1
   WHERE  id NOT IN (SELECT id FROM Database2 WHERE snap = 1340981695)
   

3. 使用EXISTS:

   SELECT *
   FROM   Database1
   WHERE  NOT EXISTS (
            SELECT *
            FROM   Database2
            WHERE id = Database1.id AND snap = 1340981695
          )