Select TOP 1 City, Count(City) AS 'MAX_COUNT' FROM Customer Group By City Order By 'MAX_COUNT' DESC; Hope this simple query will help many one. Show activity on this post. If you are using Oracle Database, you can simply use stats_mode function this will return single value with highest occurrences.
The SQL intersect operator allows us to get common values between two tables or views. The following graphic shows what the intersect does. The set theory clearly explains what an intersect does. In mathematics, the intersection of A and B (A ∩ B) is the set that contains all elements of A that also belong to B.
SELECT
<column_name>,
COUNT(<column_name>) AS `value_occurrence`
FROM
<my_table>
GROUP BY
<column_name>
ORDER BY
`value_occurrence` DESC
LIMIT 1;
Replace <column_name>
and <my_table>
. Increase 1
if you want to see the N
most common values of the column.
Try something like:
SELECT `column`
FROM `your_table`
GROUP BY `column`
ORDER BY COUNT(*) DESC
LIMIT 1;
Let us consider table name as tblperson
and column name as city
. I want to retrieve the most repeated city from the city column:
select city,count(*) as nor from tblperson
group by city
having count(*) =(select max(nor) from
(select city,count(*) as nor from tblperson group by city) tblperson)
Here nor
is an alias name.
Below query seems to work good for me in SQL Server database:
select column, COUNT(column) AS MOST_FREQUENT
from TABLE_NAME
GROUP BY column
ORDER BY COUNT(column) DESC
Result:
column MOST_FREQUENT
item1 highest count
item2 second highest
item3 third higest
..
..
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