Find Distinct Values & then Group by and count in mySQL

Question

I have a table which has some duplicate values for the same date

MyTime   |  Mrc  |  Num
21-Dec   |  ABC  |  200
21-Dec   |  ABC  |  200
20-Dec   |  DEF  |  300
20-Dec   |  DEF  |  300
19-Dec   |  XYZ  |  254
21-Nov   |  ABC  |  200

I want to ignore the duplicate entries and group by the DATE(Mytime), Mrc, Num ie I want to detect entries which have same day of the month, Mrc & Num appearing more than once but are not exact duplicate

So the expected output is:

MyTime |  Mrc  |  Num  | cnt
21     |  ABC  |  200  |  2

I tried something along the following lines:

SELECT DISTINCT(MyTime, Mrc, Num, Count(*) as cnt)
FROM  `My_Table`
WHERE MType != 'XX'
AND HType != 'YY'
GROUP BY Date(MyTime), Mrc, Num
ORDER BY MyTime DESC

But it is not working.

Answer 1

Use COUNT(DISTINCT MyTime) :

SELECT
    DAY(MyTime) as MyTime
    Mrc,
    Num,
    COUNT(DISTINCT MyTime) as cnt
FROM  `My_Table`
WHERE MType != 'XX'
AND HType != 'YY'
GROUP BY 1, 2, 3
ORDER BY 1 DESC

Notes:

• You want the DAY() function, not the DATE() function to extract the day part of the date
• DISTINCT is a keyword , not a function, so avoid using brackets ie use it like this: SELECT DISTINCT column1, column2, ...
• COUNT(DISTINCT expression) returns the number of different values