MySQL查询帮助

Question

SELECT username, (SUM(rating)/count(*)) as TheAverage, count(*) as TheCount 
FROM ratings 
WHERE month ='Aug' AND TheCount > 1 
GROUP BY username 
ORDER BY TheAverage DESC, TheCount DESC 

我知道这真的很接近（我认为），但它说“TheCount”不存在于WHERE子句和ORDER子句中。

表是：

ID，用户名，等级，月

而且我试图找出的平均得分为每一个用户，然后命令由平均等级和评级数量的结果。

请帮忙。

Answer 1

如果组和计数，你需要有：

SELECT username, (SUM(rating)/COUNT(*)) as TheAverage, Count(*) as TheCount 
    FROM rating 
    WHERE month='Aug' 
    GROUP BY username 
    HAVING TheCount > 1 
    ORDER BY TheAverage DESC, TheCount DESC 

Answer 2

SELECT username, (SUM(rating)/count()) as TheAverage, count() as TheCount 
FROM ratings 
WHERE month ='Aug' 
GROUP BY username 
HAVING TheCount > 1 
ORDER BY TheAverage DESC, TheCount DESC 

编辑：

好像我没仔细看不够。

我认为它现在会工作。

Answer 3

您可以使用AVG骨料：由month

SELECT username, month, AVG(rating) as TheAverage, COUNT(*) as TheCount 
FROM ratings 
WHERE month ='Aug' 
GROUP BY 
     username 
HAVING COUNT(*) > 1 
ORDER BY 
     TheAverage DESC, TheCount DESC 

分组是MySQL innesessary，因为你的month过滤，MySQL支持在GROUP BY查询的SELECT列表中选择未分组的列（返回组内的随机值）。