Question

我正在尝试在Oracle中运行此查询

SELECT A1.ID_USER,A1.CNAME,
 (SELECT CNAME 
   FROM
   (SELECT CNAME,COUNTER
    FROM
      (SELECT id_Category AS IDCATEGORT,COUNT(*) AS COUNTER
        FROM gamescateg
        WHERE id_game IN
       (SELECT ID_GAME
        FROM UserGames
        WHERE id_user=UsersGamers.ID_USER
        )
       GROUP BY id_Category
       )
    JOIN Categories 
    ON id_Category =IDCATEGORT
    ORDER BY COUNTER DESC
   )
  WHERE rownum = 1
 ) AS FAVORITECATEGORY
FROM UsersGamers A1
JOIN UsersGamers A2 ON A1.ID_USER= A2.ID_USER

但是我得到了错误

ORA-00904: "A2"."ID_USER": invalid identifier

A2.ID_USER需要做子查询为什么？怎么解决？你可以帮帮我吗？

我通过创造功能来解决这个问题

CREATE OR REPLACE FUNCTION FAVORITECATEGORYfunc( idnumber NUMBER )
  RETURN VARCHAR2
IS
  l_conc_names VARCHAR2(32676); 
BEGIN
  SELECT CNAME INTO l_conc_names 
   FROM
   (SELECT CNAME,COUNTER
    FROM
      (SELECT id_Category AS IDCATEGORT,COUNT(*) AS COUNTER
        FROM gamescateg
        WHERE id_game IN
       (SELECT ID_GAME
        FROM UserGames
        WHERE id_user=idnumber
        )
       GROUP BY id_Category
       )
    JOIN Categories 
    ON id_Category =IDCATEGORT
    ORDER BY COUNTER DESC
   )
  WHERE rownum = 1;
  RETURN l_conc_names;
END;

，查询是

select CNAME,FAVORITECATEGORYfunc(id_user) from usersgamers

感谢您的帮助：）

Answer 1

基本上，表面修复不会修复查询。所以，请解释你想要做什么。我怀疑查询比它需要的要复杂得多。而且，很难破译。

从表面上看，您的问题是使用表名而不是别名。因此，人们会认为像这样编写查询可以解决问题：

SELECT A1.ID_USER, A1.CNAME,
       (SELECT CNAME 
        FROM (SELECT CNAME, COUNTER
              FROM (SELECT id_Category AS IDCATEGORT, COUNT(*) AS COUNTER
                    FROM gamescateg
                    WHERE id_game IN (SELECT ug.ID_GAME
                                      FROM UserGames ug
                                      WHERE ug.id_user = UsersGamers.ID_USER
                                     )
                   GROUP BY id_Category
                  ) cc JOIN
                  Categories c
                  ON c.id_Category = cc.IDCATEGORT
             ORDER BY COUNTER DESC
            )
        WHERE rownum = 1
       ) AS FAVORITECATEGORY
FROM UsersGamers A1 JOIN
     UsersGamers A2
     ON A1.ID_USER = A2.ID_USER;

但是，唉，它不会。 Oracle限制了相关子查询的范围深度。所以，这并没有解决问题;你只会得到另一个错误。

此类查询可能正在做的最合理的事情是获取给定用户的最常见类别。如果是这样的话：

SELECT u.*
FROM (SELECT ugr.ID_USER, ugr.CNAME, c.cname, COUNT(*) as cnt,
             ROW_NUMBER() OVER (PARTITION BY ugr.ID_USER, ugr.CNAME
                                ORDER BY COUNT(*) DESC
                               ) as seqnum
      FROM UsersGamers ugr JOIN
           UserGames ug 
           ON ugr.ID_USER = ug.ID_USER JOIN
           Categories c
           ON c.id_Category = ug.id_Category
      GROUP BY ugr.ID_USER, ugr.CNAME
     ) u
WHERE seqnum = 1;

Answer 2

我已经扩展了您的查询以使其更具可读性并对问题进行了评论（无效的标识符）。这里有对别名A2的引用，其中A2不可能。

select A1.ID_USER, A1.CNAME,
(
    SELECT CNAME FROM 
    (
        SELECT CNAME, COUNTER 
        FROM
        (
            SELECT id_Category AS IDCATEGORT, 
            COUNT(*) AS COUNTER 
            FROM gamescateg 
            WHERE id_game IN 
            (
                SELECT ID_GAME FROM UserGames
                WHERE id_user=A2.ID_USER    /* your problem is here */
            ) 
            GROUP BY id_Category 
        )
        JOIN Categories 
        ON id_Category =IDCATEGORT
        ORDER BY COUNTER DESC
    )
    WHERE rownum = 1
) AS FAVORITECATEGORY
FROM UsersGamers A1 
JOIN UsersGamers A2
ON A1.ID_USER= A2.ID_USER

您是否考虑过重写以避免这么多嵌套查询？不是将它们包装在另一个中，而是尝试使用连接围绕喜欢的类别构建它。

Answer 3

尝试这样写，并确保获得理想的结果。

A 4 x 6 matrix that wraps around in steps of `3i` from `i` = 4
12 15 18 21 
24 27 30 33 
36 39 42 45 
48 51 54 57 
60 63 66 69 
72 75 78 81 

A 6 x 5 matrix with rows of `3i` for initial `i` in <11,13,17,19,23>
33 36 39 42 45 48 
39 42 45 48 51 54 
51 54 57 60 63 66 
57 60 63 66 69 72 
69 72 75 78 81 84 

A 5 x 1 matrix with rows of of ` 3i` for initial `i` in <7>
21 24 27 30 33

Sql子查询不工作＆＃34; ORA-00904：＆＃34; A2＆＃34;。＆＃34; ID_USER＆＃34;：无效的标识符＆＃34;

3 个答案: