views:

494

answers:

2

I have a table of values like this

978412, 400
978813, 20
978834, 50
981001, 20

As you can see the second number when added to the first is 1 number before the next in the sequence. The last number is not in the range (doesnt follow a direct sequence, as in the next value). What I need is a CTE (yes, ideally) that will output this

978412, 472
981001, 20

The first row contains the start number of the range then the sum of the nodes within. The next row is the next range which in this example is the same as the original data.

A: 

Check out this MSDN Article. It gives you a solution to your problem, if it will work for you depends on the ammount of data you have and your performance requirements for the query.

Edit:

Well using the example in the query, and going with his last solution the second way to get islands (first way resulted in an error on SQL 2005).

SELECT MIN(start) AS  startGroup, endGroup, (endgroup-min(start) +1) as NumNodes
FROM (SELECT g1.gapID AS start,
(SELECT min(g2.gapID) FROM #gaps g2 
WHERE g2.gapID >= g1.gapID and NOT EXISTS
(SELECT * FROM #gaps g3 
WHERE g3.gapID - g2.gapID = 1)) as endGroup
FROM #gaps g1) T1 GROUP BY endGroup

The thing I added is (endgroup-min(start) +1) as NumNodes. This will give you the counts.

JoshBerke
yes, I have this article - I just dont know how to do it.
Coolcoder
+2  A: 

From the article that Josh posted, here's my take (tested and working):

SELECT
    MAX(t1.gapID) as gapID,
    t2.gapID-MAX(t1.gapID)+t2.gapSize as gapSize
    -- max(t1) is the specific lower bound of t2 because of the group by.
FROM
  ( -- t1 is the lower boundary of an island.
    SELECT gapID
    FROM gaps tbl1 
    WHERE
      NOT EXISTS(
        SELECT *
        FROM gaps tbl2 
        WHERE tbl1.gapID = tbl2.gapID + tbl2.gapSize + 1
      )
  ) t1
  INNER JOIN ( -- t2 is the upper boundary of an island.
    SELECT gapID, gapSize
    FROM gaps tbl1 
    WHERE
      NOT EXISTS(
        SELECT * FROM gaps tbl2 
        WHERE tbl2.gapID = tbl1.gapID + tbl1.gapSize + 1
      )
  ) t2 ON t1.gapID <= t2.gapID -- For all t1, we get all bigger t2 and opposite.
GROUP BY t2.gapID, t2.gapSize
Loki
By the article benchmarks, it should be the fastest set based solution.
Loki
Well done, perfect.
Coolcoder