加入一个查询，返回每个客户最早的订单，并按最早的日期对表进行排序：

select o.*
from orders o 
inner join (
  select customer, min(created) created
  from orders
  group by customer
) c on c.customer = o.customer
order by c.created, o.customer, o.created

请看演示。
结果：

| created    | customer |
| ---------- | -------- |
| 2020-01-01 | xyz      |
| 2020-01-12 | xyz      |
| 2020-01-15 | xyz      |
| 2020-01-10 | abc      |
| 2020-01-19 | abc      |
| 2020-01-20 | abc      |

select * from orders order by customer desc, created_date asc;

2020-01-01  xyz 
2020-01-12  xyz 
2020-01-15  xyz 
2020-01-10  abc 
2020-01-19  abc 
2020-01-20  abc

可以在中使用窗口函数 order by . 所以：

select t.*
from t
order by min(created) over (partition by customer),
         customer,
         created;

请注意 customer 包含在最短的创建日期之后。这确保了如果两个客户具有相同的创建日期，那么他们的数据不会全部混合在一起。

你可以试试这个

select * from tablename
group by customer desc, created asc