如何在oracle中進行多對多的子查詢？

Question

我想通過以下代碼創建一個多對多的子查詢。

SELECT e.ep_suragate_pk FROM episode e 
    JOIN (
     SELECT n.name FROM actor n 
     JOIN episode_actor ea 
     ON n.act_suragate_pk = ea.act_suragate_pk 
    ) ep_act 
    ON ep_act.ep_suragate_pk = e.ep_suragate_pk; 

我得到一個ORA-00904：無效的標識符錯誤，但所有的列都存在。

 episode 
    ------- 
    ep_suragate_pk 


    episode_actor 
    ------------- 
    ep_suragate_pk 
    act_suragate_pk 


    actor 
    ----- 
    act_suragate_pk 

如果有人可以幫助我瞭解我要去哪裏錯了，我將不勝感激。謝謝。

Answer 1

你並不需要所有的這些子查詢

select e.ep_suragate_pk 
from episode e , episode_actor ea , actor n 
where ea.ref_id = e.ep_suragate_pk 
and  n.act_suragate_pk = ea.act_suragate_pk 

Answer 2

子查詢需要選擇列ea.ep_suragate_pk以解決無效標識問題。不確定是否會得到你想要的結果。

Answer 3

內部查詢

(
    SELECT n.name FROM actor n 
    JOIN episode_actor ea 
    ON n.act_suragate_pk = ea.act_suragate_pk 
) ep_act 

沒有列p_suragate_pk加盟

ep_act 
ON ep_act.ep_suragate_pk = e.ep_suragate_pk; 

，所以你可以修改這樣

SELECT e.ep_suragate_pk FROM episode e 
JOIN (
    SELECT n.name,n.act_suragate_pk FROM actor n 
    JOIN episode_actor ea 
    ON n.act_suragate_pk = ea.act_suragate_pk 
) ep_act 
ON ep_act.ep_suragate_pk = e.ep_suragate_pk; 

Answer 4

您使用的加入應選擇列。因此，嘗試：

SELECT e.ep_suragate_pk FROM episode e 
JOIN (
    SELECT n.name, ref_id FROM actor n 
    JOIN episode_actor ea 
    ON n.act_suragate_pk = ea.act_suragate_pk 
) ep_act 
ON ep_act.ref_id = e.ep_suragate_pk;