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SQL Grouped on column A, column B records with "higest count"

I would like to select items from a table:
- Grouped on column_A
- With the value of column_B that has the highest count of records

e.g.
column_A, column_B
car, red
car, red
car, green
motorbike, yellow
motorbike, blue
motorbike, orange
motorbike, orange

I would want the results
column_A, column_B
car, red
motorbike, orange

Where there are equal highest counts, I'm happy to take either value for column_B
0
Beamson
Asked:
Beamson
1 Solution
 
Jim HornMicrosoft SQL Server Developer, Architect, and AuthorCommented:
Let me know if this floats your boat.  Guessing there's a more elegant way to pull this off but this meets your requirements.
CREATE TABLE #foo (column_A VARCHAR(10), column_B VARCHAR(10))

INSERT INTO #foo (column_A, column_B) 
VALUES 
	('car', 'red'), 
	('car', 'red'), 
	('car', 'green'), 
	('motorbike', 'yellow'), 
	('motorbike', 'blue'), 
	('motorbike', 'orange'), 
	('motorbike', 'orange')

;WITH cte AS (
SELECT Column_A, Column_B, COUNT(Column_B) AS the_count
FROM #foo 
GROUP BY Column_A, Column_B), 
cte_grouped AS (
SELECT Column_A, Column_B, 
	RANK() OVER (PARTITION BY Column_A ORDER BY the_count DESC) AS rank_order
FROM cte) 
SELECT Column_A, Column_B FROM cte_grouped WHERE rank_order = 1

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0
 
BeamsonAuthor Commented:
That does it for me!

Thanks Jim
0
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