Oracle Sql query Group by Clause - sql

Question

MY_TABLE = Table with 2 columns Number, City.
Desired Output = City and count of unique Number associated to the city. Seattle, Bellevue is part of Combined. Even though there are 4 numbers associated to Seattle, Bellevue the output is 3 as there are only 3 distinct numbers - 123, 456, 786.
MY_TABLE
Number City
123 Seattle
456 Bellevue
789 LosAngeles
780 LosAngeles
123 Bellevue
786 Bellevue
Desired Output:
Combined 3
LosAngeles 2
Query so far:
SELECT NUMBER, CITY FROM MY_TABLE WHERE LOOKUP_ID=100 AND CITY IN
('Seattle', 'Bellevue', 'LosAngeles')
GROUP BY NUMBER, CITY
Would highly appreciate if anyone provides a recommendations around the same.

Answer 1

You could do something like
SELECT (case when city IN ('Seattle', 'Bellevue')
then 'Combined'
else city
end) city,
count( distinct number )
FROM my_table
WHERE lookup_id = 100
AND city IN ('Seattle', 'Bellevue', 'LosAngeles')
GROUP BY (case when city IN ('Seattle', 'Bellevue')
then 'Combined'
else city
end)
Of course, my guess is that you have some other table that tells you which CITY values need to be combined rather than having a hard-coded CASE statement.

Answer 2

with t as (
SELECT (
case when city IN ('Seattle', 'Bellevue')
then 'Combined'
else city
end
) city, number from my_table
)
select city, count(distinct number) from t
group by city
Tell please, if it was useful

Answer 3

Try this:
SELECT
(CASE CITY
WHEN 'Seattle' THEN ‘Combined’
WHEN 'Bellevue' THEN ‘Combined’
ELSE CITY
END), COUNT(*)
FROM
MY_TABLE
WHERE
LOOKUP_ID=100 AND CITY IN ('Seattle', 'Bellevue', 'LosAngeles')
GROUP BY
NUMBER,
(CASE CITY
WHEN 'Seattle' THEN ‘Combined’
WHEN 'Bellevue' THEN ‘Combined’
ELSE CITY
END)
that should do what you asked for, but I suspect that you have some other tables where you define which cities should be considered the same, in such a case you'll need to join on those tables

Answer 4

There are 3 answers already and none of them are generic for more cities.
Try this:
SELECT City, COUNT(Number) AS ExclusiveNumbers
FROM (SELECT q2.City, q2.CityNumCount, b.Number
FROM MY_Table b INNER JOIN
(SELECT c.City, MAX(NumOccurs) AS CityNumCount
FROM My_Table c INNER JOIN
(SELECT Number, COUNT(City) AS NumOccurs
FROM My_Table
GROUP BY Number) q1 ON c.Number = q1.Number
GROUP BY c.City) q2 ON b.City = q2.City) q3
WHERE CityNumCount = 1
GROUP BY City
UNION
SELECT 'Combined', COUNT(DISTINCT Number)
FROM (SELECT q2.City, q2.CityNumCount, b.Number
FROM MY_Table b INNER JOIN
(SELECT c.City, MAX(NumOccurs) AS CityNumCount
FROM My_Table c INNER JOIN
(SELECT Number, COUNT(City) AS NumOccurs
FROM My_Table
GROUP BY Number) q1 ON c.Number = q1.Number
GROUP BY c.City) q2 ON b.City = q2.City) q3
WHERE CityNumCount > 1
The top half of the union works out, for each City name that has no numbers in common with any other city, how many different numbers it has.
The bottom half works out the count of different numbers for cities that do have numbers in common with other cities. These 2 figures will always add up to the count of distinct numbers in the original table.