Question

首先，我应该说，我发现很难用短语来表达以下问题，因此可怕的问题标题。建议改进是非常受欢迎的。

现在回到真正的问题......

鉴于以下示例，客户和发票的无序数据......

修改我为以下

客户数据

customer_id   name
------------------
1             Gary
2             Jeremy
3             Marcia
4             Danielle

发票数据

invoice_id   customer_id   created_date   amount
------------------------------------------------
1            1             2008-01-01     500.00
2            1             2011-01-01     600.00
3            1             2012-01-01     100.00
4            1             2012-01-01     550.00
5            2             2008-01-01     600.00
6            2             2012-01-01     200.00
7            2             2013-01-01     1000.00
8            3             2012-01-01     300.00
9            3             2013-01-01     100.00
10           3             2009-01-01     250.00
11           4             2010-01-01     300.00
12           4             2011-01-01     700.00
13           4             2012-01-01     500.00

...如何编写查询以按以下方式返回数据...

第一行中最早的发票。如果同一年龄的发票超过1张，则那些年龄相同的发票，发票金额最大。如果超过1张相同年龄和金额的发票，则排序变得无关紧要。
第二行中与上一行相同的客户的下一个最旧的发票。再次，如果超过1个相同年龄的发票，发票金额最大。如果超过1张相同年龄和金额的发票，则排序变得无关紧要。

重复＃2，直到该客户没有更多发票为止。

不同客户的下一张最旧发票。如果超过1张相同的发票，那么，那些年龄相同的发票，发票金额最大。如果超过1张相同年龄和金额的发票，则排序变得无关紧要。

对＃4的同一客户重复＃2。

重复＃5，直到该客户没有更多发票为止。

重复＃4，＃5，＃6

因此，对于上面的样本数据，期望的结果将是......

customer_name invoice_id created_date amount -------------------------------------------------- Jeremy 5 2008-01-01 600.00 <-- this is the joint "oldest" invoice with id 1 but has a greater amount. Jeremy 6 2012-01-01 200.00 <-- this is the next "oldest" invoice for the same customer as the previous row. Jeremy 7 2013-01-01 1000.00 Gary 1 2008-01-01 500.00 <-- no more invoice for previous customer, so this is the next "oldest" invoice for a new customer Gary 2 2011-01-01 600.00 Gary 4 2012-01-01 550.00 <-- same age as inv_id 3 but larger amount Gary 3 2012-01-01 100.00 Marcia 10 2009-01-01 250.00 Marcia 8 2012-01-01 300.00 Marcia 9 2013-01-01 100.00 Danielle 11 2010-01-01 300.00 Danielle 12 2011-01-01 700.00 Danielle 13 2012-01-01 500.00

为了给出这个问题的更广泛背景，结果将用于追讨发票的支付，最旧和最“昂贵”是最高优先级，但随后也可以查看客户的所有发票组合在一起。

P.S。我正在使用MS SQL Server 2008。

Answer 1

希望这有效：）

with ordering as
(
  select
  row_number() over (order by o.created_date asc, o.amount desc) num,
  customer_id,
  customer_name
  from
  (
    select
    min(i.created_date) 
    over (partition by c.customer_id) as min_created_date,
    max(i.amount) 
    over (partition by c.customer_id, i.created_date) max_date_amount,
    c.name as customer_name,
    c.customer_id as customer_id,
    i.invoice_id,
    i.created_date,
    i.amount
    from
    invoice i
    join customer c on i.customer_id = c.customer_id
  )o
  where o.min_created_date = o.created_date
  and o.max_date_amount = o.amount
)
select
ord.customer_name,
i.invoice_id,
i.created_date,
i.amount
from
ordering ord
join invoice i on i.customer_id = ord.customer_id
order by ord.num asc, i.created_date asc, i.amount desc;

Answer 2

我只是把它放在这里作为已经接受的答案的替代方案。

SELECT temp.name,
    temp.Invoice_Id,
    temp.created_Date,
    temp.amount
FROM(
  SELECT 
    c.name,
    i.invoice_id,
    i.created_date,
    i.amount,  
    min(i.created_date) over (partition by c.customer_id) as min_created_date,
    max(i.customer_id) over (partition by i.created_Date, i.amount ) as customerId
  FROM
    Customer c
  LEFT JOIN 
    Invoice i
  on 
    c.customer_ID=i.Customer_ID
) temp

ORDER BY  temp.min_created_date, 
    temp.customerId desc, 
    temp.created_Date, 
    temp.amount desc

通过...订购查询结果？

2 个答案: