sql 日期相减datediff()返回两日期天数
定义和用法
datediff() 函数返回两个日期之间的天数。
语法
datediff(date1,date2)date1 和 date2 参数是合法的日期或日期/时间表达式。
注释：只有值的日期部分参与计算。
datediff(expr1,expr2): returns expr1 - expr2 as a value in days
实例一
select datediff('2008-12-30','2008-12-29') as diffdate
>
mysql> select datediff('1997-12-31 23:59:59','1997-12-30');
+----------------------------------------------+
| datediff('1997-12-31 23:59:59','1997-12-30') |
+----------------------------------------------+
|                                            1 |
+----------------------------------------------+
1 row in set (0.00 sec)
mysql>
mysql> select datediff('1997-11-30 23:59:59','1997-12-31');
+----------------------------------------------+
| datediff('1997-11-30 23:59:59','1997-12-31') |
+----------------------------------------------+
|                                          -31 |
+----------------------------------------------+
1 row in set (0.00 sec)
查找两字段之间的日期
mysql> create table employee(
    ->     id            int,
    ->     first_name    varchar(15),
    ->     last_name     varchar(15),
    ->     start_date    date,
    ->     end_date      date,
    ->     salary        float(8,2),
    ->     city          varchar(10),
    ->     description   varchar(15)
    -> );
query ok, 0 rows affected (0.03 sec)
mysql>
mysql>
mysql> insert into employee(id,first_name, last_name, start_date, end_date,   salary,  city,       description)
    ->              values (1,'jason',    'martin',  '19960725',  '20060725', 1234.56, 'toronto',  'programmer');
query ok, 1 row affected (0.00 sec)
mysql>
mysql> insert into employee(id,first_name, last_name, start_date, end_date,   salary,  city,       description)
    ->               values(2,'alison',   'mathews',  '19760321', '19860221', 6661.78, 'vancouver','tester');
query ok, 1 row affected (0.00 sec)
mysql>
mysql> insert into employee(id,first_name, last_name, start_date, end_date,   salary,  city,       description)
    ->               values(3,'james',    'smith',    '19781212', '19900315', 6544.78, 'vancouver','tester');
query ok, 1 row affected (0.00 sec)
mysql>
mysql> insert into employee(id,first_name, last_name, start_date, end_date,   salary,  city,       description)
    ->               values(4,'celia',    'rice',     '19821024', '19990421', 2344.78, 'vancouver','manager');
query ok, 1 row affected (0.00 sec)
mysql>
mysql> insert into employee(id,first_name, last_name, start_date, end_date,   salary,  city,       description)
    ->               values(5,'robert',   'black',    '19840115', '19980808', 2334.78, 'vancouver','tester');
query ok, 1 row affected (0.00 sec)
mysql>
mysql> insert into employee(id,first_name, last_name, start_date, end_date,   salary,  city,       description)
    ->               values(6,'linda',    'green',    '19870730', '19960104', 4322.78,'new york',  'tester');
query ok, 1 row affected (0.00 sec)
mysql>
mysql> insert into employee(id,first_name, last_name, start_date, end_date,   salary,  city,       description)
    ->               values(7,'david',    'larry',    '19901231', '19980212', 7897.78,'new york',  'manager');
query ok, 1 row affected (0.00 sec)
mysql>
mysql> insert into employee(id,first_name, last_name, start_date, end_date,   salary,  city,       description)
    ->               values(8,'james',    'cat',     '19960917',  '20020415', 1232.78,'vancouver', 'tester');
query ok, 1 row affected (0.00 sec)
mysql>
mysql> * from employee;
+------+------------+-----------+------------+------------+---------+-----------+-------------+
| id   | first_name | last_name | start_date | end_date   | salary  | city      | description |
+------+------------+-----------+------------+------------+---------+-----------+-------------+
|    1 | jason      | martin    | 1996-07-25 | 2006-07-25 | 1234.56 | toronto   | programmer  |
|    2 | alison     | mathews   | 1976-03-21 | 1986-02-21 | 6661.78 | vancouver | tester      |
|    3 | james      | smith     | 1978-12-12 | 1990-03-15 | 6544.78 | vancouver | tester      |
|    4 | celia      | rice      | 1982-10-24 | 1999-04-21 | 2344.78 | vancouver | manager     |
|    5 | robert     | black     | 1984-01-15 | 1998-08-08 | 2334.78 | vancouver | tester      |
|    6 | linda      | green     | 1987-07-30 | 1996-01-04 | 4322.78 | new york  | tester      |
|    7 | david      | larry     | 1990-12-31 | 1998-02-12 | 7897.78 | new york  | manager     |
|    8 | james      | cat       | 1996-09-17 | 2002-04-15 | 1232.78 | vancouver | tester      |
+------+------------+-----------+------------+------------+---------+-----------+-------------+
8 rows in set (0.00 sec)
mysql>
mysql>
mysql>
mysql> select *
    -> from employee
    -> where (datediff(curdate(),start_date)/365.25) >55;
empty set (0.00 sec)