Come posso creare una query ricorsiva in MSSQL 2005?

Question

Diciamo che ho la seguente tabella:

CustomerID ParentID Name
========== ======== ====
1          null     John
2          1        James
3          2        Jenna
4          3        Jennifer
5          3        Peter
6          5        Alice
7          5        Steve
8          1        Larry 

Voglio recuperare in una query tutti i discendenti di James (Jenna, Jennifer, Peter, Alice, Steve). Grazie, Pablo.

Answer 1

Su SQL Server 2005 è possibile utilizzare CTE (espressioni comuni di tabelle) :

with Hierachy(CustomerID, ParentID, Name, Level)
as
(
select CustomerID, ParentID, Name, 0 as Level
    from Customers c
    where c.CustomerID = 2 -- insert parameter here
    union all
    select c.CustomerID, c.ParentID, c.Name, ch.Level + 1
    from Customers c
    inner join Hierachy ch
    on c.ParentId = ch.CustomerID
)
select CustomerID, ParentID, Name
from Hierachy
where Level > 0

Answer 2

Per il bottom up usa la risposta di mathieu con una piccola modifica:

with Hierachy(CustomerID, ParentID, Name, Level)
as
(
select CustomerID, ParentID, Name, 0 as Level
    from Customers c
    where c.CustomerID = 2 -- insert parameter here
    union all
    select c.CustomerID, c.ParentID, c.Name, ch.Level + 1
    from Customers c
    inner join Hierachy ch

    -- EDITED HERE --
    on ch.ParentId = c.CustomerID
    ----------------- 

)
select CustomerID, ParentID, Name
from Hierachy
where Level > 0

Answer 3

Non è possibile eseguire la ricorsione in SQL senza stored procedure. Il modo per risolverlo è usare Nested Sets, fondamentalmente modellano un albero in SQL come set.

Si noti che ciò richiederà una modifica al modello di dati corrente o eventualmente capire come creare una vista sul modello originale.

Esempio Postgresql (usando pochissime estensioni postgresql, solo SERIAL e ON COMMIT DROP, la maggior parte dei RDBMS avrà funzionalità simili):

Configurazione:

CREATE TABLE objects(
    id SERIAL PRIMARY KEY,
    name TEXT,
    lft INT,
    rgt INT
);

INSERT INTO objects(name, lft, rgt) VALUES('The root of the tree', 1, 2);

Aggiunta di un figlio:

START TRANSACTION;

-- postgresql doesn't support variables so we create a temporary table that 
-- gets deleted after the transaction has finished.

CREATE TEMP TABLE left_tmp(
    lft INT
) ON COMMIT DROP; -- not standard sql

-- store the left of the parent for later use
INSERT INTO left_tmp (lft) VALUES((SELECT lft FROM objects WHERE name = 'The parent of the newly inserted node'));

-- move all the children already in the set to the right
-- to make room for the new child
UPDATE objects SET rgt = rgt + 2 WHERE rgt > (SELECT lft FROM left_tmp LIMIT 1);
UPDATE objects SET lft = lft + 2 WHERE lft > (SELECT lft FROM left_tmp LIMIT 1);

-- insert the new child
INSERT INTO objects(name, lft, rgt) VALUES(
    'The name of the newly inserted node', 
    (SELECT lft + 1 FROM left_tmp LIMIT 1), 
    (SELECT lft + 2 FROM left_tmp LIMIT 1)
);

COMMIT;

Visualizza una traccia dal basso verso l'alto:

SELECT
    parent.id, parent.lft
FROM
    objects AS current_node
INNER JOIN
    objects AS parent
ON
    current_node.lft BETWEEN parent.lft AND parent.rgt
WHERE
    current_node.name = 'The name of the deepest child'
ORDER BY
    parent.lft;

Visualizza l'intero albero:

SELECT
    REPEAT('   ', CAST((COUNT(parent.id) - 1) AS INT)) || '- ' || current_node.name AS indented_name
FROM
    objects current_node
INNER JOIN
    objects parent
ON
    current_node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY
    current_node.name,
    current_node.lft
ORDER BY
    current_node.lft;

Seleziona tutto in basso da un certo elemento dell'albero:

SELECT
    current_node.name AS node_name
FROM
    objects current_node
INNER JOIN
    objects parent
ON
    current_node.lft BETWEEN parent.lft AND parent.rgt
AND
    parent.name = 'child'
GROUP BY
    current_node.name,
    current_node.lft
ORDER BY
    current_node.lft;

Answer 4

A meno che non mi manchi qualcosa, la ricorsione non è necessaria ...

SELECT d.NAME FROM Customers As d
INNER JOIN Customers As p ON p.CustomerID = d.ParentID
WHERE p.Name = 'James'