（SQL）如何使用相同的ID对值进行平均

Question

我有这样的SQL代码：

SELECT ID_song as ID, rate
FROM Songs
    JOIN Rating USING (ID_song)
order by ID_song;

输出是：

  ID ¦ rate  
 ----+------ 
  27 ¦ 10    
  28 ¦ 1     
  31 ¦ 4     
  31 ¦ 1     
  41 ¦ 7     
  41 ¦ 3     
  44 ¦ 1     
  44 ¦ 7     

我希望重复ID的平均费率总是有一个唯一ID，其平均值为rate，如下所示：

输出是：

  ID ¦ rate  
 ----+------ 
  27 ¦ 10    
  28 ¦ 1     
  31 ¦ 2.5   
  41 ¦ 5     
  44 ¦ 4       

Answer 1

查询

SELECT ID_song as ID, AVG(rate)
FROM Songs
    JOIN Rating USING (ID_song)
GROUP BY ID_song
order by ID_song;

AVG()和Group By

的示例

SELECT ID,AVG(rate) FROM Table1
GROUP BY ID

<强>输出

ID  AVG(rate)
27  10
28  1
31  2.5
41  5
44  4

<强>演示

  

http://sqlfiddle.com/#!9/c17ddc/1

Answer 2

使用avg和group by - 给定using子句我假设是oracle。

SELECT ID_song as ID, avg(rate)
FROM Songs
JOIN Rating USING (ID_song)
group by ID_song
order by ID_song;

理想情况下，使用正常的ansi语法进行连接，而不是使用语法。

Answer 3

要计算平均值，请使用AVG（） 并且一定要分组。

SELECT ID_song as ID, AVG(rate)
FROM Songs
    JOIN Rating USING (ID_song)
GROUP BY ID_song
order by ID_song;

Answer 4

＆LT;

Select Distinct(Id_song) as ID, avg(rate)
from songs,rating
where song.Id_song = rating.Id_song
order by song.Id_song

＆GT;

（如果你可以分享这两个表结构，那会更有帮助。我通常不会使用连接。这种方式对我来说更容易。只是我的观点）