11

很抱歉，这篇文章很长，但我在下面提供了复制和粘贴示例数据以及可能的解决方案。问题的相关部分在帖子的上部（水平规则上方）。

我有下表

 Dt          customer_id  buy_time     money_spent
 -------------------------------------------------
 2000-01-04  100          11:00:00.00  2
 2000-01-05  100          16:00:00.00  1
 2000-01-10  100          13:00:00.00  4
 2000-01-10  100          14:00:00.00  3
 2000-01-04  200          09:00:00.00  10
 2000-01-06  200          10:00:00.00  11
 2000-01-06  200          11:00:00.00  5
 2000-01-10  200          08:00:00.00  20

并想要一个查询来获得这个结果集

 Dt          Dt_next     customer_id  buy_time     money_spent
 -------------------------------------------------------------
 2000-01-04  2000-01-05  100          11:00:00.00  2
 2000-01-05  2000-01-10  100          16:00:00.00  1
 2000-01-10  NULL        100          13:00:00.00  4
 2000-01-10  NULL        100          14:00:00.00  3
 2000-01-04  2000-01-06  200          09:00:00.00  10
 2000-01-06  2000-01-10  200          10:00:00.00  11
 2000-01-06  2000-01-10  200          11:00:00.00  5
 2000-01-10  NULL        200          08:00:00.00  20

即：我希望每个客户 ( customer_id) 和每一天 ( Dt) 第二天同一客户访问过 ( Dt_next)。

我已经有一个查询给出了后者的结果集（数据和查询包含在水平规则下方）。但是，它涉及一个left outer join和两个dense_rank聚合函数。这种方法对我来说似乎有点笨拙，我认为应该有更好的解决方案。任何指向替代解决方案的指针都非常感谢！谢谢！

顺便说一句：我使用的是 SQL Server 11，该表有 >>1m 个条目。

我的查询：

 select
   customer_table.Dt
   ,customer_table_lead.Dt as Dt_next
   ,customer_table.customer_id
   ,customer_table.buy_time
   ,customer_table.money_spent
 from
 (
   select 
     #customer_data.*
     ,dense_rank() over (partition by customer_id order by customer_id asc, Dt asc) as Dt_int
   from #customer_data
 ) as customer_table
 left outer join
 (
   select distinct
     #customer_data.Dt
     ,#customer_data.customer_id
     ,dense_rank() over (partition by customer_id order by customer_id asc, Dt asc)-1 as Dt_int
   from #customer_data
 ) as customer_table_lead
 on
 (
   customer_table.Dt_int=customer_table_lead.Dt_int
   and customer_table.customer_id=customer_table_lead.customer_id
 )

样本数据：

 create table #customer_data (
   Dt date not null,
   customer_id int not null,
   buy_time time(2) not null,
   money_spent float not null
 );

 insert into #customer_data values ('2000-01-04',100,'11:00:00',2);
 insert into #customer_data values ('2000-01-05',100,'16:00:00',1);
 insert into #customer_data values ('2000-01-10',100,'13:00:00',4);
 insert into #customer_data values ('2000-01-10',100,'14:00:00',3);

 insert into #customer_data values ('2000-01-04',200,'09:00:00',10);
 insert into #customer_data values ('2000-01-06',200,'10:00:00',11);
 insert into #customer_data values ('2000-01-06',200,'11:00:00',5);
 insert into #customer_data values ('2000-01-10',200,'08:00:00',20);
4

1 回答 1

18

试试这个查询：

select cd.Dt
    , t.Dt_next
    , cd.customer_id
    , cd.buy_time
    , cd.money_spent
from (
    select Dt
        , LEAD(Dt) OVER (PARTITION BY customer_id ORDER BY Dt) AS Dt_next
        , customer_id
    from (
        select distinct Dt, customer_id
        from #customer_data
    ) t
) t
inner join #customer_data cd on t.customer_id = cd.customer_id and t.Dt = cd.Dt

为什么字段money_spent有浮点类型？您可能在计算时遇到问题。将其转换为十进制类型。

于 2013-09-20T12:19:11.160 回答