累计每月总计和Postgresql

我试图计算dellstore2数据库的累计用户数.看看这里和其他论坛的答案,我用过这个

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

这回来了

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

看起来很好,看看前几个项目.但是当我跑的时候

select count(distinct(customerid)) from orders

对于整个事情,我明白了

8996

这与第一个输出11,681中的最后一项不一致.我猜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？

解决方法

您可以使用如下子查询,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

我认为这可以按要求工作.

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

编辑

如果你仍然需要这两种方法(即不同的和总的运行)你可以使用这个：

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