SELECT SUM(COUNT() - 1) FROM STATION GROUP BY CITY HAVING COUNT() > 1;

Best Answer is: SELECT COUNT (CITY) - COUNT(DISTINCT CITY) FROM STATION;

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SELECT COUNT(CITY) - COUNT(DISTINCT CITY) FROM STATION

SELECT COUNT(CITY) - COUNT(DISTINCT CITY) FROM STATION AS DIFFERENCE;

Hello here is the code to solve the problem

there are two ways to compute the difference between: 1. total number of CITY entries (including duplicates) 2. number of distinct CITY entries.

Assuming the table is named STATION and the column is CITY First method: 1. SQL to get the difference directly

SELECT
  COUNT(CITY) - COUNT(DISTINCT CITY) AS city_duplicate_diff
FROM STATION;

Second method: 2.If you want to see the two counts separately (total vs distinct) and compute the difference in a single result:

SELECT
  total_city AS total_city_entries,
  distinct_city AS distinct_city_entries,
  total_city - distinct_city AS city_duplicate_diff
FROM (
  SELECT
    COUNT(CITY) AS total_city,
    COUNT(DISTINCT CITY) AS distinct_city
  FROM STATION
) AS t;