How to restrict/eliminate some records that meet the condition? (right syntax)

Question

The question is How many interviews were there in 1995 and 1996. You must use the IN keyword for your comparison. Also there is a function called YEAR() that will return the year portion of a date.

TABLE:

mysql> SELECT interviewdate "Date"
    -> FROM interview;    

+------------+
| Date       |
+------------+
| 1995-06-01 |
| 1995-06-01 |
| 1995-06-30 |
| 1995-06-30 |
| 1995-07-01 |
| 1995-08-01 |
| 1995-08-01 |
| 1995-08-02 |
| 1995-12-01 |
| 1995-12-02 |
| 1995-12-04 |
| 1996-01-21 |
| 1996-02-01 |
| 1996-02-02 |
| 1996-07-01 |
| 1996-07-01 |
| 1996-08-01 |
| 1996-08-08 |
| 1996-08-11 |
| 1997-01-01 |
| 1997-01-01 |
| 1997-01-31 |
| 1997-02-01 |
| 1997-03-24 |
| 1997-03-31 |
| 1997-04-20 |
| 1997-04-22 |
| 1997-05-01 |
+------------+
28 rows in set (0.00 sec)

OUTPUT:

The condition is to count only those with the Date of 1995 and 1996 into a new table AS "Count(*)". Does anyone know how to do this? I am new to mySQL and trying to understand the syntax. I tried something like this:

SELECT interviewdate, COUNT(interviewdate) AS "COUNT(*)"
FROM interview
GROUP BY interviewdate
HAVING COUNT(interviewdate) == 1995 AND 1996;


+----------+
| COUNT(*) |
+----------+
|       19 |
+----------+
1 row in set (0.00 sec)

Answer 1

This should work:

select count(*)
from interview
where year(interviewdate) in (1995,1996)

• SQL Fiddle Demo

And if you need to insert the results into a new table, then use create table as -- more fiddle.

Answer 2

Try:

SELECT COUNT(*) AS number FROM interview WHERE (interviewdate > 1995-12-31 AND interviewdate < 1997-01-01)