Question

我有一个表，用于捕获客户购买产品的时间。它会捕获唯一的购买ID以及购买时间的时间戳。

我想查询一下今天和昨天进行了多少次购买之间的区别？

不确定如何在oracle上查询吗？

Answer 1

您可以使用条件聚合：

select sum(case when trunc(datecol) = trunc(sysdate - 1) then 1 else 0 end) as num_yesterday,
       sum(case when trunc(datecol) = trunc(sysdate) then 1 else 0 end) as num_today,
       sum(case when trunc(datecol) = trunc(sysdate) then 1
                when trunc(datecol) = trunc(sysdate - 1) then -1
           end) as diff
from t
where datecol >= trunc(sysdate - 1);

Answer 2

您可以使用“分组”功能将购买日与时间戳信息分组，并计算购买ID。

select trunc(purchase_ts) Day, count(purchase_id) Count
from purchase
group by trunc(purchase_ts)
order by 1

Answer 3

在该列上使用TRUNC将阻止Oracle在该列上使用索引（相反，您将需要一个单独的基于函数的索引）；而是使用CASE语句测试日期是否在一天的开始和第二天的开始之间，然后COUNT在这些范围之间的值：

SELECT COUNT(
         CASE
         WHEN TRUNC( SYSDATE ) - INTERVAL '1' DAY <= your_date_column
         AND  your_date_coumn < TRUNC( SYSDATE )
         THEN 1
         END
       ) AS count_for_yesterday,
       COUNT(
         CASE
         WHEN TRUNC( SYSDATE ) <= your_date_column
         AND  your_date_coumn < TRUNC( SYSDATE ) + INTERVAL '1' DAY
         THEN 1
         END
       ) AS count_for_today
FROM   your_table
WHERE  TRUNC( SYSDATE ) - INTERVAL '1' DAY <= your_date_column
AND    your_date_coumn < TRUNC( SYSDATE ) + INTERVAL '1' DAY

Oracle SQL-如何检索今天与昨天之间的ID计数差异

3 个答案: