我们如何从MySQL表中找到年龄大于30岁的雇员,并提供该表上的唯一生日?
为了理解这个概念,我们使用表'emp_tbl'中的数据,如下所示:
mysql> Select * from emp_tbl; +--------+------------+ | Name | DOB | +--------+------------+ | Gaurav | 1984-01-17 | | Gaurav | 1990-01-17 | | Rahul | 1980-05-22 | | Gurdas | 1981-05-25 | | Naveen | 1991-04-25 | | Sohan | 1987-12-26 | +--------+------------+ 6 rows in set (0.00 sec) mysql> SELECT Name,SYSDATE(),DOB,DATEDIFF(SYSDATE(),DOB)/365 AS AGE from emp_tbl WHERE(DATEDIFF(SYSDATE(), DOB)/365)>30; +--------+---------------------+------------+---------+ | Name | SYSDATE() | DOB | AGE | +--------+---------------------+------------+---------+ | Gaurav | 2017-12-26 22:33:24 | 1984-01-17 | 33.9644 | | Rahul | 2017-12-26 22:33:24 | 1980-05-22 | 37.6219 | | Gurdas | 2017-12-26 22:33:24 | 1981-05-25 | 36.6137 | | Sohan | 2017-12-26 22:33:24 | 1987-12-26 | 30.0219 | +--------+---------------------+------------+---------+ 4 rows in set (0.10 sec)