Mysql，SubQuery问题

Question

Select rating_id, average_rating

From (Select rating_id, avg(rating_num) as average_rating
    from ratings
    group by rating_id
    having count(*) > 50)

HAVING average_rating > 4 ;

运行查询后，出现错误

  

每个派生表都必须有自己的别名

我知道这里的部分有效：

Select rating_id, avg(rating_num) as average_rating
        from ratings
        group by rating_id
        having count(*) > 50

我在这个子查询中做错了什么？我搜索，搜索和搜索但无法找到错误，无论我在哪里纠正，我仍然会收到错误

Answer 1

如错误消息所示，您需要为子查询添加别名：

SELECT rating_id, average_rating
FROM (
    SELECT
        rating_id,
        AVG(rating_num) AS average_rating
    FROM ratings
    GROUP BY rating_id
    HAVING COUNT(*) > 50
) AS some_alias
WHERE average_rating > 4 

some_alias可以是任何东西 - 或者是子查询的描述性名称，或者因为您永远不需要按名称引用子查询，所以您可以使用非描述性名称，例如T1（然后是T2，T3）等等，如果你有其他子查询）。

您也可以在外部查询中使用WHERE而不是HAVING。

Answer 2

Select rating_id, average_rating

From (Select rating_id, avg(rating_num) as average_rating
    from ratings
    group by rating_id
    having count(*) > 50) a

HAVING average_rating > 4 ;

请注意表别名“a”

Answer 3

在子查询之后加上“SomeAlias”：

Select rating_id, average_rating

From (Select rating_id, avg(rating_num) as average_rating
    from ratings
    group by rating_id
    having count(*) > 50) as A

HAVING average_rating > 4 ;