根据每个名称最旧的时间戳选择行

Question

我想根据每个名称最旧的时间戳选择行。我提到这个solution但问题是我的表没有id列。我试图添加一个临时自动增量id列，但不知道如何去做。基本上当时间戳相同时，我想为给定名称选择任何一个可用记录。有人可以帮助我在这里。

mysql> select * from tblemployee order by name; 
+------+-------------+---------------------+ 
| name | day_of_week | signin_date   | 
+------+-------------+---------------------+ 
| bob | wednesday | 2017-08-11 08:11:30 | 
| bob | thursday | 2017-06-11 11:21:30 | << same timestamp 
| bob | saturday | 2017-08-28 09:01:30 | 
| bob | wednesday | 2017-08-11 08:11:30 | 
| bob | monday  | 2017-06-11 11:21:30 | << same timestamp 
| tom | wednesday | 2017-08-28 23:01:20 | 
| tom | sunday  | 2017-08-29 09:01:30 | 
+------+-------------+---------------------+ 

我预期的结果会是这样的

对于“嗵”

+------+-------------+---------------------+ 
| name | day_of_week | signin_date   | 
+------+-------------+---------------------+ 
| tom | wednesday | 2017-08-28 23:01:20 | 

为“鲍勃”因为有两个记录有相同的时间戳，我可以挑选任何一个不管day_of_week。

+------+-------------+---------------------+ 
| name | day_of_week | signin_date   | 
+------+-------------+---------------------+ 
| bob | thursday | 2017-06-11 11:21:30 | 

        (or) 

| bob | monday  | 2017-06-11 11:21:30 | 

Answer 1

您可以使用的分钟（signin_date）组值联接如：

select distinct a.name, a.day_of_week , a.signin_date   
    from tblemployee a 
    inner join (
    select name, min(signin_date) as min_date 
    from tblemployee 
    group by name 
) t on t.min_date = a.signin_date and a.name = t.name 

，如果你有ambiguos行作为鲍勃分钟（signin_date）我与两个不同的日期，然后你应该得到这一个使用

select a.name, min(a.day_of_week) , a.signin_date   
    from tblemployee a 
    inner join (
    select name, min(signin_date) as min_date 
    from tblemployee 
    group by name 
) t on t.min_date = a.signin_date and a.name = t.name 
    group by a.name,a.signin_date