“CROSS APPLY”和“OUTER APPLY”

时间:2017-03-04 06:58:56

标签: sql oracle join oracle12c

我在Oracle-12c中有一个类似于accountspostscomments的典型论坛的模式。我正在编写一个查询来获取...

  1. 一位用户
  2. 所有该用户的帖子
  3. 对每个帖子的评论
  4. 和每条评论的作者。
  5. 查询如下所示:

    select "accounts".*, "p".*, "c".*, "author".*
    from "accounts"
    cross apply (
        select * from "posts"
        where "posts"."author_id" = "accounts"."id"
    ) "p"
    cross apply (
        select * from "comments"
        where "comments"."post_id" = "p"."id"
    ) "c"
    left join "accounts" "author" on "author"."id" = "c"."author_id"
    where "accounts"."id" = 1
    

    此查询按预期工作。我正在使用CROSS APPLY而不是典型的JOIN,因为我将稍后添加OFFSETFETCH以进行分页上。但是,问题是CROSS APPLY省略了没有评论的帖子,这是我不想要的。我希望保持结果中的帖子,即使他们没有评论。

    所以我尝试将CROSS APPLY更改为OUTER APPLY

    select "accounts".*, "p".*, "c".*, "author".*
    from "accounts"
    outer apply (
        select * from "posts"
        where "posts"."author_id" = "accounts"."id"
    ) "p"
    outer apply (
        select * from "comments"
        where "comments"."post_id" = "p"."id"
    ) "c"
    left join "accounts" "author" on "author"."id" = "c"."author_id"
    where "accounts"."id" = 1
    

    但现在我收到了这个错误:

    ORA-00904: "p"."id": invalid identifier
    00904. 00000 -  "%s: invalid identifier"
    *Cause:    
    *Action:
    Error at Line: 9 Column: 34
    

    出于某种原因,我的第二个OUTER APPLY加入是抱怨我从第一个结果中引用"p"."id"。但是当我使用CROSS APPLY时它很好。

    发生了什么事?为什么这些行为之间存在这种差异?

    编辑:OUTER APPLY在这个基本示例中似乎没有必要。这是从一个更复杂的场景中提炼出来的,我必须坚持认为OUTER APPLY确实是必要的,但是这个场景的细节与我所问的实际问题无关 - 这是关于这种看似无证的行为差异在CROSSOUTER APPLY之间。

    编辑:

    Oracle版本:数据库12c标准版12.1.0.2.0版 - 64位生产

    客户端: Oracle SQL Developer版本4.2.0.16.356

    服务器uname -a - Linux ubuntu-1gb-sfo2-01 4.4.0-64-generic #85-Ubuntu SMP Mon Feb 20 11:50:30 UTC 2017 x86_64 x86_64 x86_64 GNU/Linux

    的输出

    DDL: link

2 个答案:

答案 0 :(得分:10)

CodeFuller's回答,我只想补充一点(A)有一个可用于此bug的补丁和(B)有一个可在12.1.0.2中运行的解决方法,但我不知道它是否符合您的目的。

解决方法是基本嵌套连接,如下所示:

SELECT accounts.*,
       p.*,
       author.*
FROM   accounts
       CROSS APPLY (SELECT posts.id,
                           posts.author_id,
                           posts.text,
                           comments.comment_author_id,
                           comments.comment_text
                    FROM   posts
                           OUTER APPLY (SELECT comments.author_id comment_author_id,
                                               comments.text comment_text
                                        FROM   comments
                                        WHERE  comments.post_id = posts.id) comments
                    WHERE  posts.author_id = accounts.id) p
       LEFT JOIN accounts author
         ON author.id = p.comment_author_id
WHERE  accounts.id = 1;


ID   NAME      ID_1 AUTHOR_ID TEXT                                              COMMENT_AUTHOR_ID COMMENT_TEXT                            ID_2  NAME_1                               
---- --------- ---- --------- ------------------------------------------------- ----------------- --------------------------------------- ----- ------------------- 
   1 Fred         1         1 Fred wrote this and it has comments                               3 This is Helen's comment on Fred's post      3 Helen                                
   1 Fred         2         1 Fred wrote this and it does not have any comments
-------- End of Data --------
2 row(s) fetched

参考:用于解决方法的表DDL

CREATE TABLE accounts
(
  id     NUMBER PRIMARY KEY,
  name   VARCHAR2 (30)
);

CREATE TABLE posts
(
  id          NUMBER PRIMARY KEY,
  author_id   NUMBER,
  text        VARCHAR2 (240)
);

CREATE TABLE comments
(
  id          NUMBER PRIMARY KEY,
  post_id     NUMBER,
  author_id   NUMBER,
  text        VARCHAR2 (240)
);

INSERT INTO accounts (id, name)
VALUES (1, 'Fred');

INSERT INTO accounts (id, name)
VALUES (2, 'Mary');

INSERT INTO accounts (id, name)
VALUES (3, 'Helen');

INSERT INTO accounts (id, name)
VALUES (4, 'Iqbal');

INSERT INTO posts (id, author_id, text)
VALUES (1, 1, 'Fred wrote this and it has comments');

INSERT INTO posts (id, author_id, text)
VALUES (2, 1, 'Fred wrote this and it does not have any comments');

INSERT INTO posts (id, author_id, text)
VALUES (3, 4, 'Iqbal wrote this and it does not have any comments');

INSERT INTO comments (id,
                      post_id,
                      author_id,
                      text)
VALUES (1,
        1,
        3,
        'This is Helen''s comment on Fred''s post');

答案 1 :(得分:6)

您的查询没有任何问题。您遇到了错误20356733/21547130,这是在12.1.0.2中引入的,如here所述。为了克服它,请使用12.1.0.2之前的版本或应用最新更新(修复的链接线程声明在12.1.0.2.160419补丁集更新中可用)。

此答案主要由Matthew McPeakMartin Smith找到。我刚刚按照下面的描述进行了首次尝试,发现该问题在Oracle 12.1.0.1上无法重现。

第一次回答尝试:

我已经使用您的架构创建了测试数据库,这两个查询对我来说都很好。第一个不返回没有评论的帖子,第二个返回所有帐户帖子,没有任何ORA-00904错误。我在Oracle 12c上做过这个测试。

继续提问:

  1. 尝试从帖子中复制/粘贴并执行第二个查询。有时候一些令人讨厌的错别字会潜入查询中。来自您帖子的确切查询可以按预期为我工作。

  2. 如果您仍然遇到同样的错误,请提供您用于创建accountspostscomments表格的DDL。

  3. 请指定您正在使用的SQL客户端。错误肯定是在服务器端,但在这种奇怪的情况下,每一个细节都可以产生影响。

  4. 我的测试数据库:

    CREATE TABLE "accounts"
    (
        "id" NUMBER(11) NOT NULL,
        "name" NVARCHAR2(256),
        CONSTRAINT ACCOUNTS_PK PRIMARY KEY ("id")
    )
    /
    
    CREATE TABLE "posts"
    (
        "id" NUMBER(11) NOT NULL,
        "author_id" NUMBER(11) NOT NULL,
        "post_text" NVARCHAR2(1024),
        CONSTRAINT POSTS_PK PRIMARY KEY ("id"),
        CONSTRAINT POST_ACCOUNT_FK FOREIGN KEY ("author_id") REFERENCES "accounts" ("id") ON DELETE CASCADE
    )
    /
    
    CREATE TABLE "comments"
    (
        "id" NUMBER(11) NOT NULL,
        "author_id" NUMBER(11) NOT NULL,
        "post_id" NUMBER(11) NOT NULL,
        "comment_text" NVARCHAR2(1024),
        CONSTRAINT COMMENTS_PK PRIMARY KEY ("id"),
        CONSTRAINT COMMENT_ACCOUNT_FK FOREIGN KEY ("author_id") REFERENCES "accounts" ("id") ON DELETE CASCADE,
        CONSTRAINT COMMENT_POST_FK FOREIGN KEY  ("post_id") REFERENCES "posts" ("id") ON DELETE CASCADE
    )
    /
    
    INSERT INTO "accounts"("id", "name") VALUES(1, 'testuser')
    /
    INSERT INTO "posts"("id", "author_id", "post_text") VALUES(1, 1, 'First test post')
    /
    INSERT INTO "posts"("id", "author_id", "post_text") VALUES(2, 1, 'Second test post')
    /
    INSERT INTO "comments"("id", "author_id", "post_id", "comment_text") VALUES(1, 1, 2, 'It is a very cool post')
    /
    COMMIT