具有窗口功能的高级SQL

Question

我有表a（维度表）和表B（事实表）存储交易购物者的历史记录。

表a：为唯一组合（重复的第2列，第3列，第4列中的任何一个将具有相同的购物者ID）创建的购物ID（代理键）

表b是交易数据。

我正在尝试确定每周的新客户和回头客，预期产出如下。

我正在考虑遵循SQL语句

为回头客选择COUNT（*）OVER（按shopperid，星期日期划分）作为total_new_shopperid， 为了确定在相同加入条件下的新客户（即唯一客户），我被困在窗口功能上。

谢谢

山姆

Answer 1

您可以如下使用DENSE_RANK分析函数和聚合函数：

SELECT WEEK_DATE, 
       COUNT(DISTINCT CASE WHEN DR = 1 THEN SHOPPER_ID END) AS TOTAL_NEW_CUSTOMER,
       SUM(CASE WHEN DR = 1 THEN AMOUNT END) AS TOTAL_NEW_CUSTOMER_AMT,
       COUNT(DISTINCT CASE WHEN DR > 1 THEN SHOPPER_ID END) AS TOTAL_REPEATED_CUSTOMER,
       SUM(CASE WHEN DR > 1 THEN AMOUNT END) AS TOTAL_REPEATED_CUSTOMER_AMT 
  FROM
      (
        select T.*, 
               DENSE_RANK() OVER (PARTITION BY SHOPPER_ID ORDER BY WEEK_DATE) AS DR
          FROM YOUR_TABLE T);
GROUP BY WEEK_DATE;

干杯！

Answer 2

Tejash的回答很好（我对此表示反对）。

但是，Oracle在聚合方面非常有效，因此两个聚合级别可能具有更好的性能（取决于数据）：

select week_date,
       sum(case when min_week_date = week_date then 1 else 0 end) as new_shoppers,
       sum(case when min_week_date = week_date then amount else 0 end) as new_shopper_amount,
       sum(case when min_week_date > week_date then 1 else 0 end) as returning_shoppers,
       sum(case when min_week_date > week_date then amount else 0 end) as returning_amount
from (select shopper_id, week_date,
             sum(amount) as amount,
             min(week_date) over (partition by shopper_id) as min_week_date
      from t
      group by shopper_id, week_date
     ) sw
group by week_date
order by week_date;

注意：如果这样做有更好的性能，则可能是由于取消了count(distinct)。