SQL -- 我如何写一个 select 语句来查找连续5天下订单的客户

Question

orderId、orderDate、customerId 是此处的相关字段。

一个客户可能有 1 个以上的连续 5 天。

我希望输出看起来像这样

customerID  startDate       endDate      numDays 
1           2020/01/01      2020/01/05    5
1           2020/10/1       2020/10/10   10
101         2020/04/10      2020/04/15    6

到目前为止，这是我拥有的：

    ;
with t1 as (

    select distinct o.idcustomer,orderdate, dateadd(dd,1,orderdate) nextOrderDate, 1 as tday, orderstatus
    from orders o
        join customers c on c.idcustomer=o.idcustomer
    where orderstatus in (3,4) and c.customertype=0
), t2 as (
    select * from t1
    union all
    select o2.idcustomer, o2.orderdate, dateadd(dd,1,o2.orderdate), o.tday+1, o2.orderstatus
    from t1 o2
        join t2 o on o2.idcustomer=o.idcustomer and o2.orderdate=o.nextOrderDate and o2.orderstatus in (3,4)
)

--select idcustomer, max(tday) DaysInARow, min(orderDate) StartDate, max(orderdate)  endDate
select idcustomer, dateadd(dd,-5,min(orderdate)) firstOrderDate, max(orderdate) lastOrderDate
from t2 
where tday>=5
group by idcustomer, tday
order by idcustomer

Answer 1

这是一个 gaps-and-islands 问题，您希望将客户有订单的连续几天组合在一起。

如果客户每天最多有一个订单，您可以根据递增序列使用日期算法来构建组。假设您是运行 SQL 服务器，如您当前查询的语法所示：

select customer_id, min(orderdate) startdate, max(orderdate) enddate, count(*) cnt
from (
    select c.customerid, o.orderdate, 
        row_number() over(partition by customerid order by o.orderdate) rn
    from orders o
    inner join customers c on c.idcustomer = o.idcustomer
    where o.orderstatus in (3, 4) and c.customertype = 0
) t
group by c.customer_id, dateadd(day, -rn, o.orderdate)

如果您只想显示 5 天以上的连胜，只需添加一个 having 子句：

having count(*) >= 5

而且，如果您只想让每位顾客获得最大的连续上垒次数（最少长度为 5）：

select *
from (
    select customer_id, min(orderdate) startdate, max(orderdate) enddate, count(*) cnt,
        rank() over(partition by customer_id order by count(*) desc) rn2
    from (
        select c.customerid, o.orderdate, 
            row_number() over(partition by customerid order by o.orderdate) rn
        from orders o
        inner join customers c on c.idcustomer = o.idcustomer
        where o.orderstatus in (3, 4) and c.customertype = 0
    ) t
    group by c.customer_id, dateadd(day, -rn, o.orderdate)
    having count(*) >= 5
) t
where rn2 = 1

如果有重复的(customerid, orderdate)，那么我们用dense_rank()代替row_number()，用count(distinct orderdate)代替count(*)。

Answer 2

我会使用以下方法并从如下查询开始：

SQLFiddle

WITH truncatedQueries as (
select distinct idcustomer, TRUNC(orderDate) as orderDate from orders 
),
rawData as (
select a.idcustomer,a.orderDate as order1, b.orderDate as order2, b.orderDate-a.orderDate  from truncatedQueries a inner join 
truncatedQueries b on 
a.idcustomer = b.idcustomer AND a.orderDate < b.orderDate  AND
 b.orderDate-a.orderDate =1
),
intervals as (
select idCustomer as CustomerId, min(order1) as StartDate, max(order2) as EndDate from rawData
group by idCustomer order by idCustomer
)
select CustomerId,StartDate, EndDate, endDate - StartDate as numDays  from intervals where endDate - StartDate >=5;

算法如下：

我从日期中截断时间，然后执行 distinct 查询。通过这样做，我在每个日期每个客户的每个订单中获得 1 个条目。
然后我对这些结果进行 self-join，并寻找连续的几天。我通过仅保留（同一客户的）2 个订单之间相差 1 天的行来找到连续的天数。
然后我得到如下结果：

ID    ORDERDATE   ORDERDATE   DIFFERENCE
    1 22-APR-11   23-APR-11   1
    1 23-APR-11   24-APR-11   1
    2 22-APR-11   23-APR-11   1

然后我们有了基本数据，但我们还缺少两件事：

A) 间隔的开始 B) 间隔结束 C) 它们之间的差异（天数）

为此，我使用了 min 和 max 函数。如果我们从第一个 orderDate 列获得 min，从第二个 orderDate 列获得 max，我们就会得到结果。最后一部分是做减法并在必要时检查 >=5。

Answer 3

谢谢大家！

这是满足我需要的代码：

    ;
with t1 as (

    select distinct o.idcustomer,orderdate
    from orders o
        join customers c on c.idcustomer=o.idcustomer
    where orderstatus in (3,4) and c.customertype=0
)
select idCustomer, min(orderdate) startdate, max(orderdate) enddate, count(*) cnt
from (
    select idcustomer, orderdate, 
        row_number() over(partition by idcustomer order by orderdate) rn
    from t1
) t
group by idcustomer, dateadd(day, -rn, orderdate)
having count(*) >= 5 
order by idcustomer, cnt

CTE 为我提供了唯一的 idCustomer 和 orderdate（消除了同一天的多个订单），从那里，我使用了@GMB (https://whosebug.com/users/10676716/gmb) 的第一个示例来创建输出。

使用 row_number 计算日期分组的答案非常酷。

SQL -- 我如何写一个 select 语句来查找连续5天下订单的客户

SQL -- How do I write a select statement that finds customers who placed orders on 5 consecutive days

sql

sql-server

datetime

count

gaps-and-islands