根據每個名稱最舊的時間戳選擇行

Question

我想根據每個名稱最舊的時間戳選擇行。我提到這個solution但問題是我的表沒有id列。我試圖添加一個臨時自動增量id列，但不知道如何去做。基本上當時間戳相同時，我想爲給定名稱選擇任何一個可用記錄。有人可以幫助我在這裏。

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 | 
+------+-------------+---------------------+ 

我預期的結果會是這樣的

對於「嗵」

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

爲「鮑勃」因爲有兩個記錄有相同的時間戳，我可以挑選任何一個不管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

您可以使用的分鐘（signin_date）組值聯接如：

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行作爲鮑勃分鐘（signin_date）我與兩個不同的日期，然後你應該得到這一個使用

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