one-to-many query selecting all parents and single top child for each parent

Question

There are two SQL tables:

Parents:
+--+---------+
|id|   text  |
+--+---------+
| 1|  Blah   |
| 2|  Blah2  |
| 3|  Blah3  |
+--+---------+

Childs
+--+------+-------+
|id|parent|feature|
+--+------+-------+
| 1|   1  |  123  |
| 2|   1  |   35  |
| 3|   2  |   15  |
+--+------+-------+

I want to select with single query every row from Parents table and for each one single row from Childs table with relation "parent"-"id" value and the greatest "feature" column value. In this example result should be:

+----+------+----+--------+---------+
|p.id|p.text|c.id|c.parent|c.feature|
+----+------+----+--------+---------+
|  1 | Blah |  1 |    1   |    123  |
|  2 | Blah2|  3 |    2   |    15   |
|  3 | Blah3|null|   null |   null  |
+----+------+----+--------+---------+

Where p = Parent table and c = Child table

I tried to LEFT OUTER JOIN and GROUP BY but MSSQL Express told me that query with GROUP BY require Aggregate functions on every non-Groupped fields. And I do not want to Group them all, but rather select top row (with custom ordering).

I am totally out of ideas...

Solution 1

select p.id, p.text, c.id, c.parent, c.feature
from Parents p
left join (select c1.id, c1.parent, c1.feature
             from Childs c1
             join (select p1.id, max(c2.feature) maxFeature
                     from Parents p1
                left join Childs c2 on p1.id = c2.parent
            group by p1.id) cf on c1.parent = cf.id 
                              and c1.feature = cf.maxFeature) c
on p.id = c.parent

Solution 2

Using CTE (SQL Server 2005+):

WITH max_feature AS (
   SELECT c.id,
          c.parent,
          MAX(c.feature) 'feature'
     FROM CHILD c
 GROUP BY c.id, c.parent)
   SELECT p.id,
          p.text,
          mf.id,
          mf.parent,
          mf.feature
     FROM PARENT p
LEFT JOIN max_feature mf ON mf.parent = p.id

Non CTE equivalent:

   SELECT p.id,
          p.text,
          mf.id,
          mf.parent,
          mf.feature
     FROM PARENT p
LEFT JOIN (SELECT c.id,
                  c.parent,
                  MAX(c.feature) 'feature'
             FROM CHILD c
         GROUP BY c.id, c.parent) mf ON mf.parent = p.id

Your question lacks details for handling tie breakers (when 2+ CHILD.id values have the same feature value). Agent_9191's answer uses TOP 1, but that will take the first that is returned & not necessarily the one you want.

Solution 3

This should work:

SELECT p.id, p.text, c.id, c.parent,c.feature
FROM parent p
 LEFT OUTER JOIN (SELECT TOP 1 child.id,
                               child.parent,
                               MAX(child.feature)
                  FROM child
                  WHERE child.parent = p.id
                  GROUP BY child.id, child.parent
                  ) c ON p.id = c.parent

Solution 4

manji's query does not handle tie-breakers for max feature. Here is my method, which I've tested:

;WITH WithClause AS (SELECT p.id, p.text, 
        (SELECT TOP 1 c.id from childs c 
            where c.parent = p.id order by c.feature desc) 
        AS BestChildID
    FROM Parents p) 
SELECT WithClause.id, WithClause.text, c.id, c.parent, c.feature
FROM WithClause 
LEFT JOIN childs c on WithClause.BestChildID = c.id

Solution 5

If you need to join different from a MAX column and any columns described in a group by closure of a nested select, you can use an APPLY function. It is a simplest solution. You also can use WITH operator. But that look harder.

SELECT p.id, p.text, CHILD_ROW.ANY_COLLUMN
FROM parent p
OUTER APPLY (SELECT TOP 1 child.ANY_COLLUMN
                  FROM child
                  WHERE child.parent = p.id
                  ORDER BY child.feature DESC 
                  ) CHILD_ROW

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Comments

[Raj More]

This will not return just the top child every time

[manji]

I think grouping by (c.id, c.parent) is like returning the original table (Childs) since a parent can't have the same child with 2 features.

[AndyH]

OMG Ponies' method also does not work, I tested, and I get too many rows with both his snippets, as manji indicated.

[Stijn de Witt]

Working code is most important of course, but some explanation of how it works would be great.