I've got a Dog table. Each dog has Breed and can have 0 to 2 photos. I need to recieve count of photos of all dogs for each breed: table with BreedId and matching PhotosCount. So result table should be:
BreedID|PhotosCount
-------------------
1 |3
-------------------
2 |1
-------------------
This should do the trick:
SELECT BreedID AS B, COUNT(Photo1) + COUNT(Photo2) AS C
FROM Dog
GROUP BY BreedID
COUNT aggregate function simply doesn't take into consideration NULL values. If, for a specific BreedID, all values of either Photo1 or Photo2 are NULL, then COUNT returns 0.
This should work in single scan:
SELECT
BreedID,
SUM(CASE WHEN Photo1 IS NOT NULL THEN 1 ELSE 0 END)
+ SUM(CASE WHEN Photo2 IS NOT NULL THEN 1 ELSE 0 END) [Count]
FROM Table
GROUP BY BreedID
Use Group By and SUM Of Photo1 and Photo2:
Note: If you wants the output for each dog you have to include DogId in group clause.
;WITH T AS
(
SELECT
BreedId,
SUM (CASE ISNULL(Photo1,0) WHEN 1 THEN 1 ELSE 0 END) AS Photo1,
SUM (CASE ISNULL(Photo2,0) WHEN 1 THEN 1 ELSE 0 END) AS Photo2
FROM TableName
Group By BreedId
)
SELECT
BreedId,
SUM(Photo1+Photo2) AS TotalPhoto
FROM T
Or Simply
SELECT
BreedId,
SUM (CASE ISNULL(Photo1,0) WHEN 1 THEN 1 ELSE 0 END + CASE ISNULL(Photo2,0) WHEN 1 THEN 1 ELSE 0 END) AS TotalPhoto
FROM TableName
Group By BreedId
COUNT and ISNULL together. NULL values will be counted in this caseGROUP BY two fields is also wrong :) You can just check the right answer above :)SELECT BreedID AS Breed, COUNT(Photo1) + COUNT(Photo2) AS #ofPhotos
FROM Dog
GROUP BY BreedID;
