Sql 为每个客户选择首次购买
我们试图在类似以下表格中为每位客户选择第一次购买:Sql 为每个客户选择首次购买,sql,sql-server,sql-server-2005,Sql,Sql Server,Sql Server 2005,我们试图在类似以下表格中为每位客户选择第一次购买: transaction_no customer_id operator_id purchase_date 20503 1 5 2012-08-24 20504 1 7 2013-10-15 20505 2 5 2013-09-05 20506
transaction_no customer_id operator_id purchase_date
20503 1 5 2012-08-24
20504 1 7 2013-10-15
20505 2 5 2013-09-05
20506 3 7 2010-09-06
20507 3 7 2012-07-30
我们试图实现的查询的预期结果是:
transaction_no customer_id operator_id first_occurence
20503 1 5 2012-08-24
20505 2 5 2013-09-05
20506 3 7 2010-09-06
我们得到的最接近的查询如下:
SELECT customer_id, MIN(purchase_date) As first_occurence
FROM Sales_Transactions_Header
GROUP BY customer_id;
结果如下:
customer_id first_occurence
1 2012-08-24
2 2013-09-05
3 2010-09-06
但是,当我们选择所需字段的其余部分时,我们显然必须将它们添加到GROUPBY子句中,这将使MIN的结果不同。我们也试图加入它,但没有取得任何进展
我们如何在不混淆聚合函数的情况下获得其余的相关值?听起来像是CTE的工作 CTE将允许您获得每个客户的最早购买日期。然后,您将其加入到关于customer_id和日期的原始表中,获得该事务的其余信息 像这样:
with first_date as(
select customer_id,
min(purchase_date) as first_purchase
from
table1
group by
customer_id
)
select
t1.transaction_no,
t1.customer_id,
t1.operator_id,
t1.purchase_date
from
table1 t1
inner join first_date
on
purchase_date = first_purchase
and t1.customer_id = first_date.customer_id
您可以简单地将提出的查询视为内部查询。这也适用于较旧版本的SQL Server(您没有指定SQL Server的版本) 您可以使用该功能来帮助您实现这一点 这是如何为您的案例做到这一点
WITH Occurences AS
(
SELECT
*,
ROW_NUMBER () OVER (PARTITION BY customer_id order by purchase_date ) AS "Occurence"
FROM Sales_Transactions_Header
)
SELECT
transaction_no,
customer_id,
operator_id,
purchase_date
FROM Occurences
WHERE Occurence = 1
下面的查询也将提供解决方案
select * from customer_sale_details
where purchase_date in (select min(purchase_date)
from customer_sale_details c1 group by c1.customer_id);
嗯…为什么是向下投票,它会准确返回请求的结果?可能是因为你应该在这里发布代码,而不仅仅是提供一个链接。在将H.*更改为H.transaction\u no,H.customer\u id,H.operator\u id后,这项功能非常有效,H.purchase\u date,因为表中有大量其他列。如果min purchase\u date与另一个客户的不同purchase\u date匹配,则实际上可能返回错误的值
select * from customer_sale_details
where purchase_date in (select min(purchase_date)
from customer_sale_details c1 group by c1.customer_id);