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SQL Puzzle | Getting group by sum and total sum in a single query
This puzzle is really cool. You have to get group by sum and total sum in a single query. Please check out the sample input and the expected output below-
Sample Input
ID | Val |
---|---|
1 | 10 |
1 | NULL |
2 | NULL |
2 | 100 |
2 | 200 |
3 | 200 |
3 | 100 |
4 | 156 |
4 | 255 |
4 | 244 |
4 | NULL |
4 | NULL |
4 | 345 |
Expected Output
ID | GroupBySum | TotalSum |
---|---|---|
1 | 10 | 1610 |
2 | 300 | 1610 |
3 | 300 | 1610 |
4 | 1000 | 1610 |
Rules/Restrictions
- The solution should be should use “SELECT” statement or a CTE.
- Add your solution(s) in the comments section or send you solution(s) to pawankkmr@gmail.com
Script | use below script to create table & insert some sample data
-- CREATE TABLE GroupBySumTotalSum ( ID INT ,VAL INT ) GO INSERT INTO GroupBySumTotalSum VALUES (1, 10), (1, NULL), (2, NULL), (2, 100), (2, 200), (3, 200), (3, 100), (4, 156), (4, 255), (4, 244), (4, NULL), (4, NULL), (4, 345) GO CREATE CLUSTERED INDEX Ix_ID ON GroupBySumTotalSum(ID) GO -- |
Solution 1
-- SELECT ID, SUM(Val) GroupBySum, SUM(SUM(Val)) OVER() TotalSum FROM GroupBySumTotalSum GROUP BY ID -- |
Solution 2
-- SELECT DISTINCT ID, SUM(Val) OVER (PARTITION BY ID) GroupBySum, SUM(Val) OVER() TotalSum FROM GroupBySumTotalSum -- |
Execution Plan / Performance Comparison
![Execution Plans - Getting group by sum and total sum in a single query [Multiple Solutions - Best One]](https://pawankkmr.files.wordpress.com/2016/07/execution-plans-getting-group-by-sum-and-total-sum-in-a-single-query-multiple-solutions-best-one-msbiskills-com.jpg?w=529&h=307)
Execution Plans – Getting group by sum and total sum in a single query [Multiple Solutions – Best One]
Clearly the first method wins here. In second query we are doing an explicit sort on all the columns we get in the select list. Now this sort will be very expensive as we cannot have indexes on the columns we are sorting.
Add a comment if you have any other or better solution in mind. I would love to learn it. We all need to learn.
Pawan Khowal
Pawan is a SQL Server Developer. If you need any help in writing code/puzzle or training please email at – pawankkmr”AT”gmail.com. Meanwhile please go throgh the top pages from his blog.
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with cte as(select id,SUM(val)as grptotal from GroupBySumTotalSum
group by id)
,cte1 as(select SUM(grptotal) as total from cte)
select *,(select total from cte1)as total from cte
LikeLiked by 1 person
Bro, you have used 2 Select here. The challenge is to use a SINGLE select.
LikeLike
with cte as(select distinct ID,sum(val)over(partition by id)as gt,val from #GroupBySumTotalSum )
select distinct ID,gt as grptotal,SUM(val)over()as total from cte
LikeLiked by 1 person
or this will also work.
select distinct ID,sum(val)over(partition by id)as gt,SUM(val)over()as total from #GroupBySumTotalSum
LikeLiked by 1 person
Yes this will work… 🙂 ,
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SELECT G1.ID, SUM(G1.VAL), (SELECT SUM(G2.VAL) FROM GroupBySumTotalSum G2 WHERE (G1.ID=G2.ID))
FROM GroupBySumTotalSum G1
GROUP BY G1.ID
LikeLike
I had posted the wrong query earlier which may not work for the given scenario. Here is the updated one, which works exactly as required.
SELECT G1.ID, SUM(G1.VAL), (SELECT SUM(G2.VAL) FROM GroupBySumTotalSum G2 WHERE (G1.ID=G2.ID))
FROM GroupBySumTotalSum G1
GROUP BY G1.ID
LikeLike
Bro, you have used 2 Select here. The challenge is to use a SINGLE select.
LikeLike
Select ID, SUM(VAL) As GroupBySum,
(Select Sum(VAL) from GroupBySumTotalSum) As TotalSum
from GroupBySumTotalSum
Group BY ID
LikeLiked by 1 person