你可以用 row_number() 若要对每个客户的订单进行排序，请筛选“附加”订单（即排名大于1的每个订单），然后按 brand_name :

select brand_name, count(*) no_additional_orders
from (
    select 
        t.*, 
        row_number() over(partition by customer_name order by order_time_stamp) rn
    from transactions t
) t
where rn > 1
group by brand_name

如果您还想考虑没有额外订单的品牌，则可以将过滤逻辑移到聚合函数：

select brand_name, sum(rn > 1) no_additional_orders
from (
    select t.*, row_number() over(partition by customer_name order by order_time_stamp) rn
    from transactions t
) t
group by brand_name

你的数据相当混乱。我想你想要的是最早的时间戳，而不是最早的订单。这是一个微妙的区别，但很重要：

select brand_name,
       sum(order_time_stamp > min_ots)
from (select t.*, min(order_time_stamp) over (partition by customer_name) as min_ots
      from t
     ) t
group by brand_name;

你可以做类似的事情 rank() 也：

select brand_name,
       sum(seqnum > 1)
from (select t.*, 
             rank() over (partition by customer_name order by order_time_stamp) as seqnum
      from t
     ) t
group by brand_name;

你想把每个人的所有顺序加起来 customer_name 每 brand_name 除了1，因为你不想在每个客户的第一个订单的总和。
您可以通过从订单总数中减去订购产品的不同客户的数量来完成，该数量等于每个客户的第一个订单数量：

select brand_name, 
       count(*) - count(distinct customer_name) additional_orders
from transactions
group by brand_name

请看演示。
结果：

> brand_name | additional_orders
> :--------- | ----------------:
> Pepsi      |                 0
> Fanta      |                 1