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

我知道这非常接近（我认为），但它说的是'TheCount'在WHERE子句和ORDER子句中不存在。

表格是：

id，用户名，评分，月份

我正在尝试计算每个用户的平均评分，然后按平均评分和评分数量排序结果。

请帮忙。

Answer 1

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

修改

似乎我看起来不够紧密。

我认为它现在会奏效。

Answer 2

如果你分组并计算，你需要：

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 3

您可以使用AVG聚合：

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

month分组MySQL，因为您的month已被过滤，而MySQL支持在SELECT列表GROUP BY中选择未分组的列{1}}查询（返回组内的随机值）。

MySQL查询帮助

3 个答案: