累积每月总计和Postgresql

小编典典

累积每月总计和Postgresql

sql

我正在尝试为dellstore2数据库累计计算用户数。在这里和其他论坛上寻找答案时，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

每个月是

看一下前几项，似乎总的来说还不错。但是当我跑步时

select count(distinct(customerid)) from orders

对于整个事情，我明白了

这与第一个输出11,681中的最后一项不一致。我猜上面的计算无法确定整个月的唯一性。什么是最快的计算方式（最好不使用自联接）？

阅读 203

2021-04-22

共1个答案

小编典典

除了直接从订单中选择之外，还可以使用如下子查询：

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

如果仍然需要这两种方法（即独立和总计），则可以使用以下方法：

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

2021-04-22