Question

我正在进行一项任务，要求我找到每个模块的资源数量的平均值。当前表格如下：

ResourceID   ModulID
   1            1
   2            7
   3            2
   4            4
   5            1
   6            1

基本上，我正在试图弄清楚如何获得平均资源数量。唯一的这里的相关测试数据是模块1，它有3个不同的资源连接到它。但我需要显示所有结果。

这是我的代码：

select avg(a.ress) GjSnitt, modulID
from 
(select count(ressursID) as ress 
 from ressursertiloppgave
 group by modulID) as a, ressursertiloppgave r
group by modulID;

显然它不起作用，但我目前正处于失去改变的地步。我非常感谢你们有任何意见。

Answer 1

这是您正在执行的查询，用稍微不那么钝的语法编写。

SELECT
  avg(a.ress) as GjSnitt
  , modulID
FROM
  (SELECT COUNT(ressursID) as ress 
   FROM ressursertiloppgave
   GROUP BY modulID) as a
CROSS JOIN ressursertiloppgave r    <--- Cross join are very very rare!
GROUP BY modulID;

你正在交叉加入表格，共计（6x6 =）36行并将其缩减为4，但由于总计数为36，结果是错误的。
这就是你永远不应该使用隐式连接的原因。

将查询重写为：

SELECT AVG(a.rcount) FROM 
  (select count(*) as rcount 
   FROM ressursertiloppgave r
   GROUP BY r.ModulID) a

如果您想要个人行数和底部的平均值，请执行以下操作：

SELECT r1.ModulID, count(*) as rcount
FROM ressursertiloppgave r1
GROUP BY r1.ModulID 
UNION ALL 
  SELECT 'avg = ', AVG(a.rcount) FROM 
  (select count(*) as rcount 
   FROM ressursertiloppgave r2
   GROUP BY r2.ModulID) a

同一查询中count（）的平均值

1 个答案: