SQL count row equal to column - mysql

Question

Here is my table:
+--------+---------------------+
| roomNo | date |
+--------+---------------------+
| 1 | 2017-05-17 16:05:00 |
| 1 | 2017-05-17 15:05:00 |
| 2 | 2019-05-20 12:30:00 |
| 2 | 2019-05-15 10:30:00 |
| 2 | 2019-05-14 08:00:00 |
+--------+---------------------+
I want to get the day where the room is used at least once and which day(s) had the most operations in it and how many times, in the current year. I don't know how to compare the dates.
The expected result would be something like :
+--------+------------+------------+
| roomNo | date | operations |
+--------+------------+------------+
| 2 | 2019-05-20 | 3 |
+--------+------------+------------+

Answer 1

We can use MySQL DATE function to lop off times from DATETIME and TIMESTAMP columns. Or we could use MySQL DATE_FORMAT function, to return just year, month day.
We can use an aggregate function like COUNT or SUM in a query with GROUP BY to get counts by room and day.
If "current year" means from Jan 1 thru Dec 31, we can use expression to derive date values of '2019-01-01' and '2020-01-01', and do a comparison of the date column to those values in the WHERE clause.
As a start, consider this:
SELECT t.roomno
, DATE(t.date) AS date_
, COUNT(*) AS cnt_
FROM mytable t
WHERE t.date >= DATE_FORMAT(NOW(),'%Y-01-01') + INTERVAL 0 YEAR
AND t.date < DATE_FORMAT(NOW(),'%Y-01-01') + INTERVAL 1 YEAR
GROUP
BY t.roomno
, DATE(t.date)
ORDER
BY t.roomno
, cnt_ DESC
If the goal is to just return one of the rooms that has the highest number of uses, we could use a LIMIT clause, and order by the highest count to lowest,
ORDER
BY cnt_ DESC
, t.roomno
LIMIT 1
If the results are more complex than that, we can omit the LIMIT clause, and use the result from that query as an inline view in an outer query.
With MySQL 8.0, we can use common table expression (CTE) and window/analytic functions, to get more elaborate results.