How to get the difference in years from two different dates?

Question

I want to get the difference in years from two different dates using MySQL database.

for example:

• 2011-07-20 - 2011-07-18 => 0 year
• 2011-07-20 - 2010-07-20 => 1 year
• 2011-06-15 - 2008-04-11 => 2 3 years
• 2011-06-11 - 2001-10-11 => 9 years

How about the SQL syntax? Is there any built in function from MySQL to produce the result?

Answer 1

Here's the expression that also caters for leap years:

YEAR(date1) - YEAR(date2) - (DATE_FORMAT(date1, '%m%d') < DATE_FORMAT(date2, '%m%d'))

This works because the expression (DATE_FORMAT(date1, '%m%d') < DATE_FORMAT(date2, '%m%d')) is true if date1 is "earlier in the year" than date2 and because in mysql, true = 1 and false = 0, so the adjustment is simply a matter of subtracting the "truth" of the comparison.

This gives the correct values for your test cases, except for test #3 - I think it should be "3" to be consistent with test #1:

create table so7749639 (date1 date, date2 date);
insert into so7749639 values
('2011-07-20', '2011-07-18'),
('2011-07-20', '2010-07-20'),
('2011-06-15', '2008-04-11'),
('2011-06-11', '2001-10-11'),
('2007-07-20', '2004-07-20');
select date1, date2,
YEAR(date1) - YEAR(date2)
    - (DATE_FORMAT(date1, '%m%d') < DATE_FORMAT(date2, '%m%d')) as diff_years
from so7749639;

Output:

+------------+------------+------------+
| date1      | date2      | diff_years |
+------------+------------+------------+
| 2011-07-20 | 2011-07-18 |          0 |
| 2011-07-20 | 2010-07-20 |          1 |
| 2011-06-15 | 2008-04-11 |          3 |
| 2011-06-11 | 2001-10-11 |          9 |
| 2007-07-20 | 2004-07-20 |          3 |
+------------+------------+------------+

See SQLFiddle

Answer 2

I like the solution by Bohemian, but what about using timestampdiff

select date1, date2,timestampdiff(YEAR,date2,date1) from so7749639

sqlfiddle

just seems easier.

Answer 3

mysql> SELECT FLOOR(DATEDIFF('2011-06-11','2001-10-11')/365);
+------------------------------------------------+
| FLOOR(DATEDIFF('2011-06-11','2001-10-11')/365) |
+------------------------------------------------+
|                                              9 |
+------------------------------------------------+
1 row in set (0.00 sec)

DATEDIFF() returns difference in days between two dates. This does not specifically take leap years into account but it may work in such cases:

mysql> SELECT FLOOR(DATEDIFF('2007-07-11','2004-07-11')/365);
+------------------------------------------------+
| FLOOR(DATEDIFF('2007-07-11','2004-07-11')/365) |
+------------------------------------------------+
|                                              3 |
+------------------------------------------------+
1 row in set (0.00 sec)

Answer 4

Simply by: SELECT TIMESTAMPDIFF(YEAR, date1, date2) AS difference FROM table.

Answer 5

you could just use

SELECT ROUND((TO_DAYS(date2) - TO_DAYS(date1)) / 365) ...

Also wrap it with ABS() if you want always a positive number, no matter which date precedes the other.

With ROUND(), 0.6 years will be considered 1 year, if instead you want to count only the full years, you can use FLOOR(). In this case 0.6 year will be considered 0 years, and 1.9 years will be considered 1 year.