SQL：连续行之间的差异

Question

Table 有 3 列：订单号、会员号、订单日期

需要拉取按天数细分的订单分布b/w按会员id连续2个订单

我的是这样的：

SELECT 
  a1.member_id,
  count(distinct a1.order_id) as num_orders, 
  a1.order_date, 
  DATEDIFF(DAY, a1.order_date, a2.order_date) as days_since_last_order
from orders as a1 
inner join orders as a2 
  on a2.member_id = a1.member_id+1;

它并没有完全帮助我，因为我需要的输出是：

Answer 1

您可以使用 lag() 获取同一客户上次订单的日期：

select o.*,
    datediff(
        order_date,
        lag(order_date) over(partition by member_id order by order_date, order_id)
    ) days_diff
from orders o

当同一日期有两行时，首先考虑最小的order_id。另请注意，我修复了您的 datediff() 语法：在 Hive 中，该函数只需要两个日期，没有单位。

我只是不明白你想要计算的逻辑 num_orders。

Answer 2

可能是这样的：

SELECT 
  a1.member_id,
  count(distinct a1.order_id) as num_orders, 
  a1.order_date, 
  DATEDIFF(DAY, a1.order_date, a2.order_date) as days_since_last_order
from orders as a1 
inner join orders as a2 
  on a2.member_id = a1.member_id
where not exists (
      select intermediate_order
      from orders as intermedite_order 
      where intermediate_order.order_date < a1.order_date and intermediate_order.order_date > a2.order_date) ;

SQL：连续行之间的差异

SQL: Difference between consecutive rows

sql

datetime

hive

window-functions

datediff