在MYSQL中获取其结果声明为比当前日期早12天的学生记录
为此，您需要比较并找到学生的当前日期和结果日期之间的差异。可以使用AND运算符以及DATEDIFF()。
让我们首先创建一个表-
mysql> create table DemoTable1547 -> ( -> StudentId int NOT NULL AUTO_INCREMENT PRIMARY KEY, -> StudentName varchar(20), -> StudentMarks int, -> StudentResultDeclareDate datetime -> );
使用插入命令在表中插入一些记录-
mysql> insert into DemoTable1547(StudentName,StudentMarks,StudentResultDeclareDate) values('Chris',56,'2019-10-13 13:00:00')
-> ;
mysql> insert into DemoTable1547(StudentName,StudentMarks,StudentResultDeclareDate) values('Bob',60,'2019-10-13 12:00:00');
mysql> insert into DemoTable1547(StudentName,StudentMarks,StudentResultDeclareDate) values('Mike',45,'2019-10-13 14:00:00');
mysql> insert into DemoTable1547(StudentName,StudentMarks,StudentResultDeclareDate) values('Carol',78,'2019-10-01 14:00:00');使用select语句显示表中的所有记录-
mysql> select * from DemoTable1547;
这将产生以下输出-
+-----------+-------------+--------------+--------------------------+ | StudentId | StudentName | StudentMarks | StudentResultDeclareDate | +-----------+-------------+--------------+--------------------------+ | 1 | Chris | 56 | 2019-10-13 13:00:00 | | 2 | Bob | 60 | 2019-10-13 12:00:00 | | 3 | Mike | 45 | 2019-10-13 14:00:00 | | 4 | Carol | 78 | 2019-10-01 14:00:00 | +-----------+-------------+--------------+--------------------------+ 4 rows in set (0.00 sec)
当前日期如下-
mysql> select curdate(); +------------+ | curdate() | +------------+ | 2019-10-13 | +------------+ 1 row in set (0.00 sec)
以下是查询以获取其结果在当前日期之前12天声明的学生记录的查询-
mysql> select * from DemoTable1547 where datediff(curdate(),StudentResultDeclareDate) >=12 and StudentMarks > 50;
这将产生以下输出-
+-----------+-------------+--------------+--------------------------+ | StudentId | StudentName | StudentMarks | StudentResultDeclareDate | +-----------+-------------+--------------+--------------------------+ | 4 | Carol | 7 8 | 2019-10-01 14:00:00 | +-----------+-------------+--------------+--------------------------+ 1 row in set (0.00 sec)