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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)