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SQL Puzzle | The Count Puzzle
In this puzzle you have to check for each Id we have to count a, b, c and d values. If for each Id these values are not present then do not include them. For more details please check the sample input and expected output.
Sample Input
Id | Vals |
---|---|
1 | a |
1 | b |
1 | c |
1 | d |
2 | a |
2 | d |
3 | d |
4 | g |
Expected Output
Id | cnts |
---|---|
1 | 4 |
2 | 2 |
3 | 1 |
4 | 0 |
Script
Use below script to create table and insert sample data into it.
-- CREATE TABLE Table1 ( Id VARCHAR(10) ,Vals VARCHAR(10) ) GO INSERT INTO Table1 VALUES (1, 'a'), (1, 'b'), (1, 'c'), (1, 'd'), (2, 'a'), (2, 'd'), (3, 'd') GO INSERT INTO Table1 VALUES (4,'g') GO SELECT * FROM Table1 GO -- |
Rules/Restrictions
The solution should be should use “SELECT” statement or “CTE”.
Add your solution(s) in the comments section or send you solution(s) to pawankkmr@gmail.com
SOLUTION 0 | USING SELECT & CASE
-- select Id,ISNULL(SUM(IIF(Vals IN ('a','b','c','d'),1,0)),0) CountVals FROM Table1 group by Id -- |
Output-1
-- Id CountVals ---------- ----------- 1 4 2 2 3 1 4 0 (4 rows affected) -- |
SOLUTION 1 | USING Outer Apply AND LEFT JOIN
-- SELECT n.Id, COUNT(t1.Vals) CountVals FROM ( SELECT * FROM (SELECT Id FROM Table1 GROUP BY Id) t1 OUTER APPLY ( VALUES ('a'),('b'),('c'),('d')) AS y(t) )n LEFT JOIN Table1 t1 ON t1.Vals = n.t AND t1.Id = n.Id GROUP BY n.Id GO -- |
Output-1
-- Id CountVals ---------- ----------- 1 4 2 2 3 1 4 0 (4 rows affected) -- |
SOLUTION 2 | USING PIVOT AND CASE
-- ;WITH CTE AS ( SELECT * , 'Yes' Ya FROM Table1 ) ,CTE1 AS ( SELECT Id, [a],[b],[c],[d] FROM CTE PIVOT ( MAX(Ya) FOR Vals IN ([a],[b],[c],[d]) )t ) SELECT Id , CASE WHEN [a] = 'Yes' THEN 1 ELSE 0 END + CASE WHEN [b] = 'Yes' THEN 1 ELSE 0 END + CASE WHEN [c] = 'Yes' THEN 1 ELSE 0 END + CASE WHEN [d] = 'Yes' THEN 1 ELSE 0 END Cnt FROM CTE1 -- |
Output-2
-- Id CountVals ---------- ----------- 1 4 2 2 3 1 4 0 (4 rows affected) -- |
SOLUTION 3 | USING MAX , CASE AND Group BY
-- SELECT Id , CASE WHEN [a] = 'Yes' THEN 1 ELSE 0 END + CASE WHEN [b] = 'Yes' THEN 1 ELSE 0 END + CASE WHEN [c] = 'Yes' THEN 1 ELSE 0 END + CASE WHEN [d] = 'Yes' THEN 1 ELSE 0 END Cnt FROM ( SELECT Id ,MAX(CASE WHEN Vals = 'a' THEN 'Yes' ELSE 'No' END) [a] ,MAX(CASE WHEN Vals = 'b' THEN 'Yes' ELSE 'No' END) [b] ,MAX(CASE WHEN Vals = 'c' THEN 'Yes' ELSE 'No' END) [c] ,MAX(CASE WHEN Vals = 'd' THEN 'Yes' ELSE 'No' END) [d] FROM Table1 GROUP BY Id )k -- |
Output-3
-- Id CountVals ---------- ----------- 1 4 2 2 3 1 4 0 (4 rows affected) -- |
SOLUTION 4 |
-- SELECT t.Id,SUM(CASE WHEN CountVals IS NULL THEN 0 ELSE 1 END) cnts from Table1 t LEFT JOIN ( SELECT n.Id, COUNT(n.Vals) CountVals FROM Table1 n WHERE vals IN ('a','b','c','d') GROUP BY n.Id )k on t.Id = k.Id GROUP BY t.Id GO -- |
Output-4
-- Id CountVals ---------- ----------- 1 4 2 2 3 1 4 0 (4 rows affected) -- |
SOLUTION 5 | Using CASE and SUM
-- SELECT Id ,SUM(CASE WHEN Vals = 'a' THEN 1 ELSE 0 END) +SUM(CASE WHEN Vals = 'b' THEN 1 ELSE 0 END) +SUM(CASE WHEN Vals = 'c' THEN 1 ELSE 0 END) +SUM(CASE WHEN Vals = 'd' THEN 1 ELSE 0 END) Cnt FROM Table1 GROUP BY Id GO -- |
Output-5
-- Id CountVals ---------- ----------- 1 4 2 2 3 1 4 0 (4 rows affected) -- |
Add a comment if you have any other solution in mind. I would love to learn it. We all need to learn.
Enjoy !!! Keep Learning
Pawan Khowal
Http://MSBISkills.com
SELECT Id,
SUM(
CASE WHEN Vals IN (‘a’,’b’,’c’,’d’) Then 1
ELSE 0 END
)Cnt
FROM Table1
GROUP BY Id
LikeLiked by 1 person
The last row 4 -> 0 is not coming. Could you please check.?
LikeLike
;with cte
as
(
select ID,
case when vals in (‘a’,’b’,’c’,’d’)then id end as cnt
from #Table1
)
select ID,COUNT(cnt) as cnt
from cte
group by ID
LikeLiked by 1 person