为此,您可以使用INTERVAL和DATE_SUB()的概念。让我们首先创建一个表-
create table DemoTable1845 ( ArrivalDate date );
使用插入命令在表中插入一些记录-
insert into DemoTable1845 values('2019-12-02'); insert into DemoTable1845 values('2019-11-18'); insert into DemoTable1845 values('2019-12-18'); insert into DemoTable1845 values('2019-12-25'); insert into DemoTable1845 values('2019-11-15');
使用select语句显示表中的所有记录-
select * from DemoTable1845;
这将产生以下输出-
+-------------+ | ArrivalDate | +-------------+ | 2019-12-02 | | 2019-11-18 | | 2019-12-18 | | 2019-12-25 | | 2019-11-15 | +-------------+ 5 rows in set (0.00 sec)
当前日期如下-
select now(); +---------------------+ | now() | +---------------------+ | 2019-12-02 20:47:05 | +---------------------+ 1 row in set (0.00 sec)
这是查询以获取今天前15天的记录-
select * from DemoTable1845 where ArrivalDate < DATE_SUB(NOW(), INTERVAL 15 DAY);
这将产生以下输出-
+-------------+ | ArrivalDate | +-------------+ | 2019-11-15 | +-------------+ 1 row in set (0.00 sec)