sin SQL查询时如何按天数不同分组？

Question

我的查询中有一个最终连接，它连接了 2 个表 t1 和 t2，

select count(t1.customerID) as count, DATEDIFF(day,t2.action_date,min(t1.ordered_date)) as diff_date 
inner join t2
on t1.EMAIL = t2.EMAIL and t1.CATEGORY_ID = t2.CATEGORY_ID  
inner join product on prod.CATEGORY_ID= t1.CATEGORY_ID
where t2.action_date <= t1.ordered_date
group by t2.action_date,t1.CATEGORY_ID

我得到如下结果

Count    diff_date
100        0
222        0
300        0
20         1
40         1

我想要的是得到这样的东西

Count      difff_date
622          0
60           1
and so on...

我不确定如何得到这样的结果？谁能帮忙？

Answer 1

您需要另一个聚合级别：

SELECT SUM(count) count, diff_date
FROM (
  SELECT COUNT(t1.customerID) count, DATEDIFF(day, t2.action_date, min(t1.ordered_date)) diff_date 
  FROM t1
  INNER JOIN t2 ON t1.EMAIL = t2.EMAIL AND t1.CATEGORY_ID = t2.CATEGORY_ID  
  INNER JOIN product p ON p.CATEGORY_ID = t1.CATEGORY_ID
  WHERE t2.action_date <= t1.ordered_date
  GROUP BY t2.action_date, t1.CATEGORY_ID
) t
GROUP BY diff_date;

或者，使用 SUM() window 函数：

SELECT DISTINCT 
       SUM(COUNT(t1.customerID)) OVER (PARTITION BY DATEDIFF(day, t2.action_date, min(t1.ordered_date))) count, 
       DATEDIFF(day, t2.action_date, min(t1.ordered_date)) diff_date 
FROM t1
INNER JOIN t2 ON t1.EMAIL = t2.EMAIL and t1.CATEGORY_ID = t2.CATEGORY_ID  
INNER JOIN product p ON p.CATEGORY_ID = t1.CATEGORY_ID
WHERE t2.action_date <= t1.ordered_date
GROUP BY t2.action_date, t1.CATEGORY_ID;

sin SQL查询时如何按天数不同分组？

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