Question

很抱歉这篇长篇文章，但我提供了副本＆amp;粘贴样本数据和下面可能的解决方案方法。问题的相关部分位于帖子的上半部分（水平线规则上方）。

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

并希望查询获得此结果集

 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）和每天（Dt）在同一客户访问过的第二天（Dt_next）。

我已经有一个查询给出后一个结果集（数据和查询包含在水平规则下面）。但是，它涉及left outer join和两个dense_rank聚合函数。这种方法对我来说似乎有些笨拙，我认为应该有更好的解决方案。 任何对替代解决方案的指示都非常感谢！谢谢！

顺便说一句：我使用的是SQL Server 11，表中有＆gt;＆gt; 1m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
 )

示例数据：

 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);

Answer 1

尝试此查询：

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具有浮点类型？您可能遇到计算问题。将其转换为十进制类型。

SQL Server：跨组（而不是组内）的超前/滞后分析功能

1 个答案: