问题描述
当我运行以下 SQL 语句时:
When I run the following SQL statement:
SELECT MAX(SUM(TIME))
FROM downloads
GROUP BY SSN
它返回客户下载的最大总和值,但是如果我尝试通过将其添加到 select 语句来找到该最大值所属的社会安全号码:
It returns the maximum sum value of downloads by a customer, however if I try to find the social security number that that max value belongs to by adding it to the select statement:
SELECT SSN, MAX(SUM(TIME))
FROM downloads
GROUP BY SSN
我收到以下错误:
不是单组群功能
我不明白为什么它会抛出这个错误.谷歌搜索提出了以下操作:
I do not understand why it is throwing this error. A google search came up with the following action:
从 SELECT 列表中删除组函数或单个列表达式,或添加包含所有列出的单个列表达式的 GROUP BY 子句
Drop either the group function or the individual column expression from the SELECT list or add a GROUP BY clause that includes all individual column expressions listed
我认为这是在说- 删除组函数使总和值无效- 删除单个列表达式 (SSN) 只会给我最大总和- 不确定第三部分.
From what I think this is saying - dropping the group function makes the sum value invalid - droping the individual column expression (SSN) will just give me the max sum - not sure about that third part.
谁能指引正确的方向?
-托梅克
这个数据库中的TIME指的是下载的次数
TIME in this database refers to the number of times downloaded
推荐答案
问题简单来说就是查询中特定 SSN 的 SUM(TIME) 是单个值,因此它反对 MAX没有意义(单个值的最大值是没有意义的).
Well the problem simply-put is that the SUM(TIME) for a specific SSN on your query is a single value, so it's objecting to MAX as it makes no sense (The maximum of a single value is meaningless).
不确定您使用的是什么 SQL 数据库服务器,但我怀疑您想要一个更像这样的查询(以 MSSQL 背景编写 - 可能需要对您正在使用的 sql 服务器进行一些翻译):
Not sure what SQL database server you're using but I suspect you want a query more like this (Written with a MSSQL background - may need some translating to the sql server you're using):
SELECT TOP 1 SSN, SUM(TIME)
FROM downloads
GROUP BY SSN
ORDER BY 2 DESC
这将为您提供总时间最长的 SSN 及其总时间.
This will give you the SSN with the highest total time and the total time for it.
编辑 - 如果你有多个相同的时间并且想要它们全部你会使用:
Edit - If you have multiple with an equal time and want them all you would use:
SELECT
SSN, SUM(TIME)
FROM downloads
GROUP BY SSN
HAVING SUM(TIME)=(SELECT MAX(SUM(TIME)) FROM downloads GROUP BY SSN))
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