Question

我在学校有一个数据库项目，我差不多完成了。我唯一需要的是每天平均电影。我有一个watchhistory，你可以找到看电影的用户。指示是你过滤掉那些每天平均有2部电影的史密斯人。

我写了以下SQL语句。但每次我都会遇到错误。有人能帮助我吗？

SQL：

SELECT
    customer_mail_address,
    COUNT(movie_id) AS AantalBekeken,
    COUNT(movie_id) / SUM(GETDATE() -
        (SELECT subscription_start FROM Customer)) AS AveragePerDay
FROM
    Watchhistory
GROUP BY
    customer_mail_address

错误：

Msg 130，Level 15，State 1，Line 1
无法对包含聚合或子查询的表达式执行聚合函数。

我尝试了不同的东西，这个查询总结了每天的电影总数。现在我需要一切的平均值，而SQL只显示每天平均有超过2部电影的cusotmer。

SELECT 
    Count(movie_id) as AantalPerDag, 
    Customer_mail_address, 
    Cast(watchhistory.watch_date as Date) as Date
FROM 
    Watchhistory
GROUP BY 
    customer_mail_address, Cast(watch_date as Date)

Answer 1

我看到的一个大问题是，您尝试使用子查询，就像它是单个值一样。子查询可能会返回许多值，除非您的系统中只有一个客户，否则它将完全执行此操作。您应该JOIN到Customer表。希望JOIN仅在WatchHistory中每行返回一个客户。如果情况并非如此，那么您将有更多的工作要做。

SELECT
    customer_mail_address,
    COUNT(movie_id) AS AantalBekeken,
    CAST(COUNT(movie_id) AS DECIMAL(10, 4)) / DATEDIFF(dy, C.subscription_start, GETDATE()) AS AveragePerDay
FROM
    WatchHistory WH
INNER JOIN Customer C ON C.customer_id = WH.customer_id  -- I'm guessing at the join criteria here since no table structures were provided
GROUP BY
    C.customer_mail_address,
    C.subscription_start
HAVING
    COUNT(movie_id) / DATEDIFF(dy, C.subscription_start, GETDATE()) <> 2

我猜测标准每天不是完全 2部电影，但要么少于2部分或超过2部分。您需要根据情况进行调整那。此外，您需要根据需要调整平均值的精度。

Answer 2

错误消息告诉您的是，您不能将SUM与COUNT一起使用。

尝试将SUM（GETDATE（） - （SELECT subscription_start FROM Customer））作为第二个聚合变量，并且

尝试使用HAVING＆amp;在查询结束时过滤以仅选择具有count / sum = 2

的用户

Answer 3

也许这就是你需要的？让我们加入两个表Watchhistory和Customers

select  customer_mail_address, 
COUNT(movie_id) AS AantalBekeken,
COUNT(movie_id) / datediff(Day, GETDATE(),Customer.subscription_start) AS AveragePerDay
from Watchhistory inner join Customer 
on Watchhistory.customer_mail_address = Customer.customer_mail_address
GROUP BY
    customer_mail_address
having AveragePerDay = 2

根据你的需要改变最后一行代码（我不明白你是否想要它进出）

Answer 4

我知道了。最后:)）

SELECT  customer_mail_address, SUM(AveragePerDay) / COUNT(customer_mail_address) AS gemiddelde
FROM            (SELECT DISTINCT customer_mail_address, COUNT(CAST(watch_date AS date)) AS AveragePerDay
                      FROM            dbo.Watchhistory
                      GROUP BY customer_mail_address, CAST(watch_date AS date)) AS d
GROUP BY customer_mail_address
HAVING        (SUM(AveragePerDay) / COUNT(customer_mail_address) >= 2

访问：每天获得普通电影有困难

4 个答案: