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プロパティを列に変換する

タイプのリストがあります：

SELECT * FROM type;

id          name
----------- --------------------------------------------------
1           person
2           other god
3           location
4           role
5           gender

そして、それぞれがタイプを持つオブジェクトのリスト：

SELECT * FROM object;

id          name                                               type_id
----------- -------------------------------------------------- -----------
1           Adam                                               1
2           Eve                                                1
3           Cain                                               1
4           Abel                                               1
5           Jeroboam                                           1
6           Zeredah                                            3

そして型名を表示するビュー：

SELECT * FROM object_view;

id          name                                               type_name
----------- -------------------------------------------------- --------------------------------------------------
1           Adam                                               person
2           Eve                                                person
3           Cain                                               person
4           Abel                                               person
5           Jeroboam                                           person
6           Zeredah                                            location

関係の種類のリスト：

SELECT * FROM relationship;

id          name
----------- --------------------------------------------------
1           has father
2           has mother
3           from

そしてオブジェクト間の関係のリスト：

SELECT * FROM object_relationship;

object_a_id relationship_id object_b_id
----------- --------------- -----------
4           1               1
3           2               2
5           3               6

これらの関係のビューと同様に：

SELECT * FROM object_relationship_view;

object_a                                           relationship                                       object_b
-------------------------------------------------- -------------------------------------------------- --------------------------------------------------
Abel                                               has father                                         Adam
Cain                                               has mother                                         Eve
Jeroboam                                           from                                               Zeredah

father、mother、およびfromの列を持つ各オブジェクトをリストしたいと思います。オブジェクトにこれらのプロパティのいずれかがない場合、列にはNULLが表示されます。したがって、結果は次のようになります。

enter image description here

うまくいくと思われるアプローチの1つを次に示します。

SELECT      id,

            object.name,

            (
                SELECT      (SELECT name FROM object WHERE id = object_b_id)
                FROM        object_relationship 
                WHERE       object.id = object_relationship.object_a_id AND 
                            object_relationship.relationship_id = (SELECT id FROM relationship WHERE name = 'has father')
            ) AS father,

            (
                SELECT      (SELECT name FROM object WHERE id = object_b_id)
                FROM        object_relationship 
                WHERE       object.id = object_relationship.object_a_id AND 
                            object_relationship.relationship_id = (SELECT id FROM relationship WHERE name = 'has mother')
            ) AS mother,

            (
                SELECT      (SELECT name FROM object WHERE id = object_b_id)
                FROM        object_relationship 
                WHERE       object.id = object_relationship.object_a_id AND 
                            object_relationship.relationship_id = (SELECT id FROM relationship WHERE name = 'from')
            ) AS [from]

FROM        object;

私の質問はこれです：これはJOINを介して行うことができますか？

このアプローチは近いです：

SELECT      object.name,
            (SELECT name FROM object WHERE id = REL_FATHER.object_b_id) AS father,
            (SELECT name FROM object WHERE id = REL_MOTHER.object_b_id) AS mother,
            (SELECT name FROM object WHERE id = REL_FROM.object_b_id)   AS [from]

FROM        object
            LEFT JOIN   object_relationship AS REL_FATHER   ON  object.id = REL_FATHER.object_a_id
            LEFT JOIN   object_relationship AS REL_MOTHER   ON  object.id = REL_MOTHER.object_a_id
            LEFT JOIN   object_relationship AS REL_FROM     ON  object.id = REL_FROM.object_a_id

WHERE       REL_FATHER.relationship_id = (SELECT id FROM relationship WHERE name = 'has father')    AND
            REL_MOTHER.relationship_id = (SELECT id FROM relationship WHERE name = 'has mother')    AND
            REL_FROM.relationship_id   = (SELECT id FROM relationship WHERE name = 'from');

このアプローチの問題は、father、mother、およびfromの値を持つオブジェクトのみがリストされることです。これらのいずれかがNULLの場合、リストされません。

したがって、たとえば、次のように追加された関係データがあるとします。

EXEC insert_object_relationship 'Abel', 'has father', 'Adam';
EXEC insert_object_relationship 'Abel', 'has mother', 'Eve';
EXEC insert_object_relationship 'Abel', 'from',       'Eden';

EXEC insert_object_relationship 'Cain', 'has father', 'Adam';
EXEC insert_object_relationship 'Cain', 'has mother', 'Eve';
EXEC insert_object_relationship 'Cain', 'from',       'Eden';

EXEC insert_object_relationship 'Jeroboam', 'from', 'Zeredah';

上記のクエリは次を返します。

enter image description here

（Zeredahはfatherとmotherの関係がないため、リストされていません。

上記のどちらよりも優れたアプローチはありますか？

上記の手法は新しいものではないと確信しています。これについて議論している参照へのポインタは大歓迎です。 （つまり、データベース理論のテキストにこの名前がありますか？）

これらのテーブルとデータを生成するために必要なすべてのコードを以下に示します。

この質問の方がスタックオーバーフローに適していると思われる場合は、お知らせください。そちらでお尋ねします。

提案をありがとう！


DROP TABLE IF EXISTS object_relationship;
DROP TABLE IF EXISTS object;
--------------------------------------------------------------------------------
DROP TABLE IF EXISTS type;

CREATE TABLE type
(
    id      INT             NOT NULL    PRIMARY KEY     IDENTITY(1, 1),
    name    nvarchar(50)    NOT NULL
);
--------------------------------------------------------------------------------
CREATE TABLE object
(
    id      INT             NOT NULL    PRIMARY KEY     IDENTITY(1, 1),
    name    nvarchar(50)    NOT NULL,
    type_id int             NOT NULL    CONSTRAINT FK_object_type FOREIGN KEY REFERENCES type(id)
);
--------------------------------------------------------------------------------
DROP TABLE IF EXISTS relationship;

CREATE TABLE relationship
(
    id      INT             NOT NULL    PRIMARY KEY     IDENTITY(1, 1),
    name    nvarchar(50)    NOT NULL
);
--------------------------------------------------------------------------------
CREATE TABLE object_relationship
(
    object_a_id     INT                 CONSTRAINT FK_object_relationship_object_object_a   FOREIGN KEY REFERENCES object(id),
    relationship_id INT                 CONSTRAINT FK_object_relationship_relationship      FOREIGN KEY REFERENCES relationship(id),
    object_b_id     INT                 CONSTRAINT FK_object_relationship_object_object_b   FOREIGN KEY REFERENCES object(id)
);
--------------------------------------------------------------------------------
DROP VIEW IF EXISTS object_view;
GO

CREATE VIEW object_view

AS

SELECT  object.id, object.name AS name, type.name AS type_name
FROM    object INNER JOIN type ON object.type_id = type.id;

GO
--------------------------------------------------------------------------------
DROP VIEW IF EXISTS object_relationship_view;
GO

CREATE VIEW object_relationship_view

AS

SELECT      A.name AS object_a, relationship.name AS relationship, B.name AS object_b
FROM        object AS A INNER JOIN object_relationship ON A.id = object_relationship.object_a_id
                        INNER JOIN relationship ON object_relationship.relationship_id = relationship.id
                        INNER JOIN object AS B ON B.id = object_relationship.object_b_id;
GO
--------------------------------------------------------------------------------
INSERT INTO type (name)
VALUES
('person'),
('other god'),
('location'),
('role'),
('gender');
DROP PROC IF EXISTS insert_object;      

GO

CREATE PROC insert_object
    @object     AS nvarchar(50),
    @type       AS nvarchar(50)

AS

INSERT INTO object (name, type_id)
VALUES
(@object, (SELECT id FROM type WHERE name = @type));

GO
--------------------------------------------------------------------------------
EXEC insert_object 'Adam',      'person';
EXEC insert_object 'Eve',       'person';
EXEC insert_object 'Cain',      'person';
EXEC insert_object 'Abel',      'person';
EXEC insert_object 'Jeroboam',  'person';
EXEC insert_object 'Zeredah',   'location';
EXEC insert_object 'Eden',  'location';
--------------------------------------------------------------------------------
INSERT INTO relationship (name)
VALUES
('has father'),
('has mother'),
('from');
--------------------------------------------------------------------------------
DROP PROC IF EXISTS insert_object_relationship;
GO

CREATE PROC insert_object_relationship
    @a  AS nvarchar(50),
    @relationship AS nvarchar(50),
    @b AS nvarchar(50)

AS

INSERT INTO object_relationship (object_a_id, relationship_id, object_b_id)
VALUES
((SELECT id FROM object WHERE name = @a), (SELECT id FROM relationship WHERE name = @relationship), (SELECT id FROM object WHERE name = @b));

GO
--------------------------------------------------------------------------------
EXEC insert_object_relationship 'Abel', 'has father', 'Adam';
EXEC insert_object_relationship 'Cain', 'has mother', 'Eve';
EXEC insert_object_relationship 'Jeroboam', 'from', 'Zeredah';
sql-servert-sqlpivot
2
dharmatech

上記のテクニックは新しいものではないと確信しています...

そうですね、あなたが探している用語は「ピボット」だと思います。これを行うには、T-SQL PIVOT 演算子を使用できます。

SELECT 
    pivot_table.aName,
    pivot_table.[has mother],
    pivot_table.[has father],
    pivot_table.[from]
FROM
(
    SELECT 
        oA.[name] AS aName, 
        oB.[name] AS bName, 
        r.[name] AS rName
    FROM dbo.[object] oA
    LEFT JOIN dbo.object_relationship ore
        ON ore.object_a_id = oA.id
    LEFT JOIN dbo.relationship r
        ON r.id = ore.relationship_id
    LEFT JOIN dbo.[object] oB
        ON ob.id = ore.object_b_id
) source_table
PIVOT
(
    MAX(bName)
    FOR rName IN ([has mother], [has father], [from])
) AS pivot_table
ORDER BY pivot_table.aName;

そして、あなたの投稿の下部にあるスクリプトで与えられたサンプルデータの結果：

screenshot of query results in SSMS

2
Josh Darnell

上記のジョシュの答えに基づいて別のアプローチがあります。

SELECT *
FROM    object_relationship_view
PIVOT
(
    MAX(object_b) FOR relationship IN ([has mother], [has father], [from])
) AS PIVOT_TABLE
ORDER BY object_a;

サブクエリの代わりに、object_relationship_viewを使用します。

この場合、すべての列にNULLを含む行が出力されないという点で、出力は少し異なります。

enter image description here

1
dharmatech