MySQL - 获取给定列值列表的最新记录

问题描述 投票:1回答:3

我有以下表结构:

请求表:

 id  insret_time             account id
 ------------------------------------
 1  2018-04-05 08:06:23       abc
 2  2018-09-03 08:14:45       abc
 3  2018-08-13 09:23:34       xyz
 4  2018-08-04 09:25:37       def
 5  2018-08-24 11:45:37       def

我需要找到帐户ID abc和def的最新记录。我不在乎xyz。

我试图使用group by和inner join方法获得结果,但是没有成功将结果限制为我关心的用户列表。请指教

更新:感谢大家的反馈。欣赏它!我需要整行作为输出。我使用id列而不是timestamp来获取最新记录,因为它自动递增这是我最终想出的,它给了我需要的输出:

select t.* FROM table t
join (select max(table.id) as maxNum from table 
where account_id in ('abc','def') group by account_id) tm on t.id = tm.maxNum;
mysql sql hibernate
3个回答
0
投票

我想这就是你要找的东西

  select account_id,max(insret_time)
  from table where account_id in ('abc', 'def')
  group by account_id

-1
投票

仅使用where获取insert_timein的最大值,仅包括'abc','def':

select t.* from tablename as t where
t.accountid in ('abc', 'def') and
t.insert_time = (select max(insert_time) from tablename where tablename.accountid = t.accountid)

-1
投票

您可以使用not in并忽略xyz记录并通过desc下订单:

select account_id,max(insert_time) from table where account_id not in ('xyz') group by account_id order by id desc

您也可以将!=运算符用于一个表达式:

select account_id,max(insert_time) from table where account_id!='xyz' group by account_id order by id desc

我希望它有帮助:)

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