我在做什麼錯在甲骨文SqlDevelopper此相關查詢

Question

我試圖建立一個查詢來選擇相關係數爲用戶確定最受好評的電影

我的表的結構是這樣的：

我的查詢是這樣的：

SELECT 
    user1, user2, 
    ((psum - (sum1 * sum2/n))/sqrt((sum1sq - pow(sum1, 2.0)/n) * (sum2sq - pow(sum2, 2.0)/n))) AS r, 
    n 
FROM 
    (SELECT 
     n1.idclient AS user1, 
     n2.idclient AS user2, 
     SUM(n1.cote) AS sum1, 
     SUM(n2.cote) AS sum2, 
     SUM(n1.cote * n1.cote) AS sum1sq, 
     SUM(n2.cote * n2.cote) AS sum2sq, 
     SUM(n1.cote * n2.cote) AS psum, 
     COUNT(*) AS n 
    FROM 
     cote AS n1 // <---------- Editor is pointing a missing parenthese here with a red line 
    LEFT JOIN 
     cote AS n2 ON n1.idfilm = n2.idfilm 
    WHERE 
     n1.idclient > n2.idclient 
    GROUP BY 
     n1.idclient, n2.idclient) AS step1 
ORDER BY 
    r DESC, n DESC 

但我相信，我不會錯過任何一個括號，我做了什麼W¯¯榮？

P.S：我不是專家，我想明白

Answer 1

Oracle不會讓你使用AS對錶的別名，只爲列別名。 The select-list syntax顯示可選的AS; the table-reference syntax沒有。

在線路你突出

FROM cote AS n1 

的AS被視爲表別名（雖然它是一個保留字，所以你不能使用它的一個別名）;那麼它看到n1並不知道它的意思。解析器似乎往往回落至猜測，任何打開的括號中應該已經關閉，馬上在這種情況下n1令牌之前，因爲這可能已經完成了它迄今（我非常簡化，當然）分析;因此不完全有用的錯誤消息。

所以，你需要刪除所有三個地方AS你使用它：

SELECT 
     user1, user2, 
     ((psum - (sum1 * sum2/n))/sqrt((sum1sq - pow(sum1, 2.0)/n) * (sum2sq - pow(sum2, 2.0)/n))) AS r, 
     n 
FROM 
     (SELECT 
       n1.idclient AS user1, 
       n2.idclient AS user2, 
       SUM(n1.cote) AS sum1, 
       SUM(n2.cote) AS sum2, 
       SUM(n1.cote * n1.cote) AS sum1sq, 
       SUM(n2.cote * n2.cote) AS sum2sq, 
       SUM(n1.cote * n2.cote) AS psum, 
       COUNT(*) AS n 
     FROM 
       cote n1 
    LEFT JOIN 
     cote n2 
    ON 
     n1.idfilm = n2.idfilm 
     WHERE 
       n1.idclient > n2.idclient 
    GROUP BY 
     n1.idclient, n2.idclient) step1 
ORDER BY 
     r DESC, 
     n DESC