Question

我正试图找出编写查询的最佳方法，这将帮助我确定本月的新客户，但还有三个月前与我们进行过交易但本月才回来的老客户。

select customer_id
 from(
   select s.customer_id,
        min(ss.tran_date) as customer_first_date,
        max(ss.tran_date) as customer_last_date
  from shipments s
  left join shipment_stops ss on ss.customer_id = s.customer_id) 
  where date_trunc('month',customer_first_date) = date_trunc('month',now()) OR ?????

任何帮助或指向正确方向的信息将不胜感激。

非常感谢您。

最好

Answer 1

您指的是最近一个月，因此请查看having子句中的最新交易，然后再查看select中的最早交易：

select s.customer_id,
       (case when min(ss.tran_date) >= date_trunc('month', now()) 
             then 'New'
             else 'Existing'
        end) as grouping
from shipments s left join
     shipment_stops ss
     on ss.customer_id = s.customer_id 
group by s.customer_id
having max(ss.tran_date) >= date_trunc('month', now())

Answer 2

使用子查询-选择3个月大的客户和当月客户，然后与当前月的3个月大客户进行左联接，如果两个客户都匹配，则为老客户，否则为新客户

with t1 as
    ( select s.* from from shipments s
      left join shipment_stops ss on ss.customer_id = s.customer_id)
    where date_trunc('month',customer_first_date)<= date_trunc('month',CURRENT_DATE - INTERVAL '3 months')
    ) , t2 as
 (
   select s.* from from shipments s
      left join shipment_stops ss on ss.customer_id = s.customer_id)
    where date_trunc('month',customer_first_date)>= date_trunc('month',now())
 ) select case when t2.customer_id=t1.customer_id then 'old customer'
   else 'new customer who comes this month' end customer_status
     left join t1
  on t2.customer_id =t1.customer_id

Answer 3

找到本月所有交易的客户（使用exists子查询）并进行一些条件汇总以区分新客户和退货客户：

SELECT
    customer_id
  , CASE
        WHEN prev_max_dt < (current_date - '3 month'::INTERVAL)
        THEN 'Returning'
        ELSE 'New'
    END
FROM (
    SELECT
        ss.customer_id
      , MAX(CASE
            WHEN ss.tran_date < date_trunc('month', current_date) THEN ss.tran_date
        END) AS prev_max_dt
    FROM shipment_stops ss
    WHERE EXISTS (
        SELECT NULL
        FROM shipment_stops
        WHERE tran_date >= date_trunc('month', current_date)
        AND customer_id = ss.customer_id
        )
    GROUP BY
        ss.customer_id
) d

查询逻辑以识别新客户和回头客

3 个答案: