mysql的，子查詢的問題

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 ; 

運行查詢後，我得到一個錯誤

每個派生的表必須有它自己的別名

我知道的部分在這裏工作：

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

我在這個子查詢中做錯了什麼？我搜查，搜查，搜查，但未能發現其中的錯誤，無論身在何處我糾正，我仍然得到錯誤

Answer 1

認沽「爲SomeAlias」子查詢後：

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 ; 

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 ; 

說明表的別名「一」

Answer 3

隨着錯誤消息指出您需要添加一個別名，那麼子查詢：

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可以是任何東西 - 要麼是子查詢的描述性名稱，要麼因爲您永遠不需要通過名稱引用子查詢，您可以使用非描述性名稱，如T1（然後是T2，T3等，如果有其他子查詢）。

此外，您可以在外部查詢中使用WHERE而不是HAVING。