MySQL 如何从MySQL表中找到年龄大于30岁的员工，只提供表中的出生日期？

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)

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