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Comment obtenir le parent donné à un enfant dans SQL SERVER 2005

Je pense que vous devriez renommer votre child_id en node, votre parent_id en child_of. Votre nom de colonne est un peu déroutant

create table stack_overflow
(
node int, child_of int
);


insert into stack_overflow(node, child_of) values
(1,0),
(2,1),
(3,2),
(4,2),
(5,3),
(6,4),
(7,0),
(8,7),
(9,8),
(10,1);

Cela fonctionne sur n'importe quel SGBDR compatible CTE :

with find_parent(parent, child_of, recentness) as
(
    select node, child_of, 0 
    from stack_overflow
    where node = 9
    union all
    select i.node, i.child_of, fp.recentness + 1
    from stack_overflow i
    join find_parent fp on i.node = fp.child_of
)
select top 1 parent from find_parent 
order by recentness desc

Sortie :

parent
7

[EDIT :plus flexible et pérenne] :

with find_parent(node_group, parent, child_of, recentness) as
(
    select node, node, child_of, 0
    from stack_overflow
    where node in (5,9)
    union all
    select fp.node_group, i.node, i.child_of, fp.recentness + 1
    from stack_overflow i
    join find_parent fp on i.node = fp.child_of
)
select q.node_group as to_find, parent as found 
from find_parent q 
join
(
    select node_group, max(recentness) as answer
    from find_parent
    group by node_group 
) as ans on q.node_group = ans.node_group and q.recentness = ans.answer 
order by to_find    

Sortie :

to_find     found
5           1
9           7

Si vous utilisez Postgres , le code ci-dessus pourrait être raccourci en :

with recursive find_parent(node_group, parent, child_of, recentness) as
(
    select node, node, child_of, 0
    from stack_overflow
    where node in (5,9)
    union all
    select fp.node_group, i.node, i.child_of, fp.recentness + 1
    from stack_overflow i
    join find_parent fp on i.node = fp.child_of
)
select distinct on (node_group) node_group as to_find, parent as found 
from find_parent 
order by to_find, recentness desc

DISTINCT SUR les rochers! :-)