sql - 带窗口函数的高级 SQL

Question

我有表 a（维度表）和表 B（事实表）存储交易购物者历史记录。

表 a：为唯一组合创建的购物 id（代理键）（第 2 列、第 3 列、第 4 列中的任何重复将具有相同的购物者 id）

表b是交易数据。

我正在尝试确定每周的新客户和重复客户，预期输出如下。

我正在考虑遵循 SQL 语句

选择 COUNT(*) OVER (PARTITION BY shopperid,weekdate) 作为重复客户的 total_new_shopperid，在相同的加入条件下识别新客户（即唯一），我被困在窗口功能上。

谢谢，

山姆

Answer 1

您可以将DENSE_RANK分析函数与聚合函数一起使用，如下所示：

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（我赞成）。

但是，Oracle 在聚合方面非常高效，因此两个级别的聚合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：如果这有更好的性能，可能是由于消除了count(distinct).