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I have a table with two id columns id1 and id2. Each row shows that 2 id values have been linked together. My actual data has both id columns as varchar(20) but I have used integers in the example below to illustrate the problem.

id1 id2
--- ---
1    2
1    3
3    4
3    5
6    7

I want to group together ids so that all linked ids are grouped together even if there is not a direct link between 2 id values.

id group_id
-- ----
1   1
2   1
3   1
4   1
5   1
6   2
7   2   

I would like to achieve this in sql without using cursors if possible.

All advice gratefully received.

Update: I thought it would be useful to add a bit more background on the data this represents. The data represents records for businesses sourced from different data sources

The main table looks like this

id business_name postal_address business_classification x y group_id

The main table can contain duplicates / similar records which need to be associated. The table with id1 and id2 columns is a result of de-duplication using fuzzy string matching (levenshtein, word matching etc) and spatial proximity matching. I think that this table is actually an undirected graph containing many disconnected sub graphs e.g. group_id 1 and 2.I want to join all ids in the same network and update the group_id in the main table.

I think that the question is now how to identify all disconnected sub graphs on a graph using SQL server Find All Disconnected Sub Graphs - Java

share|improve this question
I don't understand why 1,2,3,4,5 are in group 1, and 6,7 are in group 2. – Aaron Bertrand Mar 26 '14 at 15:45
How derp can this hierarchy go? – billinkc Mar 26 '14 at 15:59
Also, which version of SQL Server are you using (please say 2012) – billinkc Mar 26 '14 at 16:00
Is the data guaranteed to have no cycles? I.e., (1, 2), (2, 3), (3, 1) is not possible? – Jon Seigel Mar 26 '14 at 17:09
The data can have cycles unfortunately – Matt20013 Mar 26 '14 at 17:19
up vote 0 down vote accepted

Well this worked for the given sample data but not the cycles scenario / may not work for a more complex data set. Thought I would post anyway as a starter for 10:

DECLARE @t TABLE ( id1 INT, id2 INT )

    ( 1, 2 ),
    ( 1, 3 ),
    ( 3, 4 ),
    ( 3, 5 ),
    ( 6, 7 )

    --( 1, 2 ),
    --( 2, 3 ),
    --( 3, 1 )

;WITH cte AS
SELECT ROW_NUMBER() OVER( ORDER BY id1 ) group_id, id1 id 
FROM @t a
    CROSS APPLY ( SELECT TOP 1 id2 FROM @t c WHERE a.id1 = c.id1 ) d 
WHERE NOT EXISTS ( SELECT * FROM @t b WHERE a.id1 = b.id2 )
  AND a.id2 = d.id2
SELECT group_id, t.id2
FROM cte c
    INNER JOIN @t t ON = t.id1
FROM cte
share|improve this answer
This looks great. I have also looked at the cycles. The direction of id1, id2 does not matter in my data set so I can break the cycle by changing the direction using update @t set id1 = id2, id2 = id1 where id2 < id1 which then works with the cycle above. This brings back some duplicates so I changed the final select statement to SELECT distinct id1,id2 FROM cte ORDER BY 1, 2 – Matt20013 Mar 27 '14 at 10:18
That's good news Matt. I did wonder if there might be a simpler way, eg with LAG/LEAD but sometimes it's good to get something working and build on it. – wBob Mar 27 '14 at 10:33

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