11

I have the following data in my database:

Parent      Child
101         102
101         103
101         104
101         105
101         106

My parameter is 106. And using the parameter I want to retrieve all the other children under its parent which is 101. I tried using the recursive method but it didn't work given the following data. Is there another way to formulate a query?

lc.
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waine
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4 Answers4

16

Assuming you want to get siblings of the value @p0, you can use a simple self-join:

SELECT p.Child
FROM Table1 c
INNER JOIN Table1 p ON c.Parent = p.Parent
WHERE c.Child = @p0
AND p.Child <> @p0

The not-equal clause here makes sure you get siblings not including the value you searched for. Remove it as necessary.

SQL Fiddle example


Since you mention recursion though, perhaps you want the entire tree starting at the parent of the value @p0. In which case, you can use a recursive CTE:

WITH parent AS (
    SELECT Parent
    FROM Table1
    WHERE Child = @p0
), tree AS (
    SELECT x.Parent, x.Child
    FROM Table1 x
    INNER JOIN parent ON x.Parent = parent.Parent
    UNION ALL
    SELECT y.Parent, y.Child
    FROM Table1 y
    INNER JOIN tree t ON y.Parent = t.Child
)
SELECT Parent, Child
FROM tree

SQL Fiddle examples using your data and with additional data to demonstrate the recursive CTE

lc.
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7

SQL Authority has a blog with a very nice explanation of how to perform a Hierarchical Query using a Recursive CTE

http://blog.sqlauthority.com/2012/04/24/sql-server-introduction-to-hierarchical-query-using-a-recursive-cte-a-primer/

Regards

Ricardo Huertas
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-1
select child
from my_table T1
where exists (select 1 from my_table T2 where child = @parameter and T1.parent = T2.parent)
Vulcronos
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-2
WITH parent AS (
    SELECT Parent
    FROM Table1
    WHERE Child = @p0
), tree AS (
    SELECT x.Parent, x.Child
    FROM Table1 x
    INNER JOIN parent ON x.Parent = parent.Parent
    UNION ALL
    SELECT y.Parent, y.Child
    FROM Table1 y
    INNER JOIN tree t ON y.Parent = t.Child
)
SELECT Parent, Child
FROM tree
Meyer
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