我们如何从 MySQL 表中找到年龄大于 30 岁的员工，并提供表中唯一的出生日期？

mysql> Select * from emp_tbl;
+--------+------------+
| Name   | DOB        |
+--------+------------+
| Gaurav | 1984-01-17 |
| Gaurav | 1990-01-17 |
| Rahul  | 1980-05-两二 |
| Gurdas | 1981-05-二5 |
| Naveen | 1991-04-两5 |
| Sohan  | 1987-1二-两6 |
+--------+------------+
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 | 两017-1两-二6 二两:33:两4 | 1984-01-17 | 33.9644 |
| Rahul  | 两017-1两-两6 两二:33:二4 | 1980-05-两二 | 37.6两19 |
| Gurdas | 两017-1两-二6 两两:33:两4 | 1981-05-两5 | 36.6137 |
| Sohan  | 二017-1两-两6 二两:33:二4 | 1987-1二-两6 | 30.0两19 |
+--------+---------------------+------------+---------+
4 rows in set (0.10 sec)