创建sql查询以在时间范围内获取重复列值

Question

有人可以帮助我处理此查询吗？我想所有的customer_id的其中重复24小时

SELECT 
    O.Order_No, O.Customer_ID, O.DateOrdered, O.IPAddress, 
    C.FirstName, C.LastName, CD.nameoncard 
FROM 
    Order_No O 
INNER JOIN 
    CardData CD ON O.card_id = CD.id 
INNER JOIN 
    Customers C ON O.customer_id = C.customer_id 
ORDER BY 
    O.order_no desc 

增加更多的细节不止一次的结果.. 客户ID XX因此，假设订单于04/23下午2:30和2日再次订单在同一天04/23下午5:30与同一客户Id xx一起发出。

我想查询返回我的客户ID XX

感谢

Answer 1

select Customer_ID, CAST(DateOrdered as Date) DateOrdered, count(*) QTDE 
from Order_No 
group by Customer_ID, CAST(DateOrdered as Date) 
having count(*) > 1 

要得到谁拥有第一一个后发出订单的客户，那么你可以使用下面的查询：

select distinct A.Customer_ID 
from Order_No A 
inner join (select Customer_ID, min(DateOrdered) DateOrdered from Order_No group by Customer_ID) B 
    on A.Customer_ID = B.Customer_ID 
    and A.DateOrdered - B.DateOrdered <= 1 
    and A.DateOrdered > B.DateOrdered 

SQL Fiddle

要获得有不止一个数量级的期间发行不超过24h或等于随时为所有客户

select distinct A.Customer_ID 
from Order_No A 
inner join Order_No B 
    on A.Customer_ID = B.Customer_ID 
    and A.DateOrdered > B.DateOrdered 
    and A.DateOrdered - B.DateOrdered <= 1 

SQL Fiddle

Answer 2

自连接：

SELECT distinct O.Customer_ID 
FROM 
Order_No O 
inner join Order_No o2 
on o.customerID = o2.customerID 
    and datediff(hour, o.DateOrdered, o2.DateOrdered) between 0 and 24 
and o.Order_No <> o2.Order_No 

这将返回所有CUSTOMER_IDS在任何24小时的时间内都有多个订单。

编辑添加匹配记录不应该是相同的记录的连接标准。应该返回同时放置两个不同订单的客户，而不是放置一个订单的客户。