如何查找一个值是否存在于 VARRAY 中

Question

我在 table 中创建了一个 VARRAY（下图）我想查询标题是否具有特定主题，例如。显示 'Action' 场比赛。我不确定该怎么做...

CREATE OR REPLACE TYPE Theme_Game AS OBJECT
(Theme VARCHAR(20));
/ 
CREATE OR REPLACE TYPE Theme_Type AS VARRAY(3) OF Theme_Game;
/
CREATE OR REPLACE TYPE Game_Type AS OBJECT
(Title VARCHAR2(50),
GameTheme Theme_Type);
/
CREATE TABLE Game_Table of Game_Type
/
INSERT INTO Game_Table
VALUES('Star Wars' ,(Theme_Type(Theme_Game('Action'), Theme_Game('FPS'))))
/

Answer 1

您需要使用 table() 函数在 FROM 子句中公开嵌套的 table。然后您可以引用集合的属性：

SQL> select g.title
  2  from game_table g
  3       , table(g.gametheme) gt
  4  where gt.theme = 'Action';

TITLE
--------------------------------------------------
Star Wars

SQL>

"what if I then needed to retrieve rows with multiple Themes i.e Action, FPS?"

对于笨拙的解决方案表示歉意，但我现在需要去上班。稍后我可能会 post 一个更优雅的解决方案。

SQL> select * from game_table
  2  /

TITLE
--------------------------------------------------
GAMETHEME(THEME)
--------------------------------------------------------------------------------
Star Wars
THEME_TYPE(THEME_GAME('Action'), THEME_GAME('FPS'))

Uncharted 3
THEME_TYPE(THEME_GAME('Action'), THEME_GAME('Puzzle'))

Commander Cody
THEME_TYPE(THEME_GAME('Fun'), THEME_GAME('Puzzle'))


SQL> select g.title
  2  from game_table g
  3       , table(g.gametheme) gt
  4       , table(g.gametheme) gt1
  5  where gt.theme = 'Action'
  6  and gt1.theme = 'FPS' ;

TITLE
--------------------------------------------------
Star Wars

SQL>

此替代方法不适用于您当前的类型，因为 VARRAY 不支持 member of。但是如果集合是嵌套的 Table.

 select g.title
  from game_table g
  where  'Action' member of g.gametheme
  and 'FPS' member of g.gametheme

Answer 2

对于多个主题，您可以这样做

select g.Title
from game_table g, table(g.gameTheme) t
where t.Theme in ('FPS','Action')
group by g.Title having count(0) = 2;

这也可以让你做一些事情，比如获得恰好 n 次匹配、至少 n 次匹配、最多 n 次匹配的标题...

Answer 3

您可以使用一个集合，然后使用 SUBMULTISET 运算符比较多个项目：

SQL Fiddle

Oracle 11g R2 架构设置:

CREATE OR REPLACE TYPE Theme_Game AS OBJECT
(Theme VARCHAR(20));
/ 
CREATE OR REPLACE TYPE Theme_Type AS TABLE OF Theme_Game;
/
CREATE OR REPLACE TYPE Game_Type AS OBJECT(
  Title VARCHAR2(50),
  GameTheme Theme_Type
);
/
CREATE TABLE Game_Table of Game_Type
NESTED TABLE GameTheme STORE AS GameTheme_Tab
/
INSERT INTO Game_Table
VALUES('Star Wars' ,(Theme_Type(Theme_Game('Action'), Theme_Game('FPS'))))
/

查询 1:

SELECT *
FROM   game_table
WHERE  Theme_Type(Theme_Game('Action'), Theme_Game('FPS'))
       SUBMULTISET OF GameTheme

Results:

|     TITLE |                                             GAMETHEME |
|-----------|-------------------------------------------------------|
| Star Wars | oracle.sql.STRUCT@67e8dc0f,oracle.sql.STRUCT@795b6d4c |

但是，为什么要使用只有一个 VARCHAR2 属性的 Theme_Game 对象？您可以只使用 VARRAY(3) OF VARCHAR2(20) 或 TABLE OF VARCHAR2(20) 而无需中间对象：

SQL Fiddle

Oracle 11g R2 架构设置:

CREATE OR REPLACE TYPE Varchar20List AS TABLE OF VARCHAR2(20);
/
CREATE OR REPLACE TYPE Game_Type AS OBJECT(
  Title VARCHAR2(50),
  GameTheme Varchar20List
);
/
CREATE TABLE Game_Table of Game_Type
NESTED TABLE GameTheme STORE AS GameTheme_Tab
/
INSERT INTO Game_Table
VALUES('Star Wars' , Varchar20List('Action', 'FPS'))
/

查询 1:

SELECT *
FROM   game_table
WHERE  Varchar20List('Action','FPS')
       SUBMULTISET OF GameTheme

Results:

|     TITLE |  GAMETHEME |
|-----------|------------|
| Star Wars | Action,FPS |

如果你想用 VARRAYs 做，那么：

SQL Fiddle

Oracle 11g R2 架构设置:

CREATE OR REPLACE TYPE Varchar20List AS VARRAY(3) OF VARCHAR2(20);
/
CREATE OR REPLACE TYPE Game_Type AS OBJECT(
  Title VARCHAR2(50),
  GameTheme Varchar20List
);
/
CREATE TABLE Game_Table of Game_Type
/
INSERT INTO Game_Table
VALUES('Star Wars' , Varchar20List('Action', 'FPS'))
/

查询 1:

SELECT *
FROM   game_table g
WHERE  2 >= ( SELECT COUNT(*)
              FROM   TABLE( g.GameTheme ) a
                     INNER JOIN
                     TABLE( Varchar20List( 'Action', 'FPS' ) ) b
                     ON ( a.COLUMN_VALUE = b.COLUMN_VALUE )
             )

Results:

|     TITLE |  GAMETHEME |
|-----------|------------|
| Star Wars | Action,FPS |

或：

查询 2:

SELECT *
FROM   game_table g
WHERE  2 >= ( SELECT COUNT(*)
              FROM   TABLE( g.GameTheme ) a
              WHERE  a.COLUMN_VALUE IN ( 'Action', 'FPS' )
             )

Results:

|     TITLE |  GAMETHEME |
|-----------|------------|
| Star Wars | Action,FPS |

如何查找一个值是否存在于 VARRAY 中

How to find if a value exists within a VARRAY

sql

oracle

user-defined-types