5

我正在尝试为 dellstore2 数据库累积计算用户数。在这里查看答案和其他论坛，我使用了这个

select 
date_trunc('month',orderdate),
sum(count(distinct(customerid)))
   over (order by date_trunc('month',orderdate))
from orders group by date_trunc('month',orderdate)

这返回

2004-01-01 00:00:00.0   979
2004-02-01 00:00:00.0   1,952
2004-03-01 00:00:00.0   2,922
2004-04-01 00:00:00.0   3,898
2004-05-01 00:00:00.0   4,873
2004-06-01 00:00:00.0   5,846
2004-07-01 00:00:00.0   6,827
2004-08-01 00:00:00.0   7,799
2004-09-01 00:00:00.0   8,765
2004-10-01 00:00:00.0   9,745
2004-11-01 00:00:00.0   10,710
2004-12-01 00:00:00.0   11,681

每个月都是

979
973
970
976
975
973
981
972
966
980
965
971

看着前几项，似乎总体上还不错。但是当我跑

select count(distinct(customerid)) from orders

对于整个事情，我得到

8996

这与第一个输出 11,681 中的最后一项不一致。我想上面的计算不能确定几个月的唯一性。这种计算的最快方法是什么，最好不使用自连接？

4

1 回答 1

7

您可以使用这样的子查询，而不是直接从订单中选择：

SELECT  OrderDate,
        SUM(COUNT(DISTINCT customerid)) OVER (ORDER BY OrderDate)
FROM    (   SELECT  CustomerID, 
                    DATE_TRUNC('MONTH', MIN(OrderDate)) AS OrderDate
            FROM    Orders
            GROUP BY CustomerID
        ) AS Orders
GROUP BY OrderDate

我认为这将按要求工作。

http://sqlfiddle.com/#!1/7a8cc/1

编辑

如果您仍然需要这两种方法（即不同的和运行总计），您可以使用这个：

SELECT  OrderDate,
        COUNT(DISTINCT CustomerID) AS MonthTotal,
        SUM(COUNT(DISTINCT customerid)) OVER (ORDER BY OrderDate) AS CumulativeTotal,
        SUM(COUNT(DISTINCT CASE WHEN OrderNumber = 1 THEN customerid END)) OVER (ORDER BY OrderDate) AS CumulativeDistinctTotal
FROM    (   SELECT  CustomerID, 
                    OrderDate,
                    ROW_NUMBER() OVER(PARTITION BY CustomerID ORDER BY OrderDate) AS OrderNumber
            FROM    Orders
        ) AS Orders
GROUP BY OrderDate

这里的例子：

http://sqlfiddle.com/#!1/7a8cc/10

于 2012-06-12T11:25:44.207 回答