如何从两个不同的日期获得多年的差异？

Question

我希望使用MySQL数据库从两个不同的日期获得差异.

例如:

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

SQL语法怎么样？MySQL有没有内置函数来产生结果？

Answer 1

这里的表达也适合闰年:

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

这工作,因为表达的(DATE_FORMAT(date1, '%m%d') < DATE_FORMAT(date2, '%m%d'))是true,如果DATE1是"今年早些时候"比date2的,并因为在MySQL中,true = 1并且false = 0,这样的调整是简单地减去比较的"真理"的问题.

这给出了测试用例的正确值,除了测试#3 - 我认为它应该是"3"以与测试#1保持一致:

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;

输出:

+------------+------------+------------+
| 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 |
+------------+------------+------------+

请参见SQLFiddle

Answer 2

我喜欢Bohemian的解决方案,但是如何使用timestampdiff

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

sqlfiddle

看起来更容易

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()返回两个日期之间的天数差异.这并没有特别考虑闰年,但在这种情况下它可能会起作用:

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

简单地说:SELECT TIMESTAMPDIFF(YEAR, date1, date2) AS difference FROM table.