Question

我要做的是一个简单的推荐者，必须采用最大加权的前40个元素的node2元素。权重的计算来自(E.WEIGHT * K.GRADE)。现在，此代码成功返回前40个元素。但是，我不希望E.NODE2返回重复项。 POSTGRE SQL允许我做SELECT DISTINCT ON (NODE2) E.NODE2 , (E.WEIGHT * K.GRADE)。我怎样才能在oracle中做同样的事情？

完整的SQL查询;

SELECT *
 FROM   (SELECT DISTINCT E.NODE2  , (E.WEIGHT * K.GRADE)
    FROM KUAISFAST K, EDGES E
    WHERE K.ID = 1 AND K.COURSE_ID = E.NODE1 AND E.NODE2 NOT IN(
        SELECT K2.COURSE_ID
        FROM KUAISFAST K2
        WHERE K2.ID = 1
        ) 
    ORDER BY( E.WEIGHT * K.GRADE ) DESC) TEMP
WHERE rownum <= 40

Answer 1

我相信你想要像

这样的东西

SELECT *
FROM   (
  SELECT 
      E.NODE2,
     (E.WEIGHT * K.GRADE),
      ROW_NUMBER() OVER (PARTITION BY E.NODE2 ORDER BY E.WEIGHT * K.GRADE DESC) R
  FROM 
      KUAISFAST K,
      EDGES E
  WHERE 
      K.ID = 1 AND 
      K.COURSE_ID = E.NODE1 AND 
      E.NODE2 NOT IN
        (  SELECT K2.COURSE_ID
           FROM KUAISFAST K2
          WHERE K2.ID = 1
        ) 
    ORDER BY (E.WEIGHT * K.GRADE) DESC
) TEMP
WHERE R=1 AND 
ROWNUM <= 40

Answer 2

在您的子选择中，我认为您需要：MAX(E.WEIGTH * K.GRADE)以便每个E.NODE2只返回一个值。

这意味着您还需要GROUP BY E.NODE2。

Answer 3

这应该可以解决你的问题，但速度很慢

SELECT * FROM
(SELECT *
 FROM   (SELECT E.NODE2 ,  max(E.WEIGHT * K.GRADE ) AS MAXDE
    FROM KUAISFAST K, EDGES E
    WHERE K.ID = 1 AND K.COURSE_ID = E.NODE1 AND E.NODE2 NOT IN(
        SELECT K2.COURSE_ID
        FROM KUAISFAST K2
        WHERE K2.ID = 1
        ) 
        GROUP BY E.NODE2 )
ORDER BY MAXDE DESC)
WHERE rownum <= 40

Oracle上的区别选择

3 个答案: