这是MySql5.1参考手册上的一个例子:表这这样的,表名为pet。
+----------+--------+---------+------+------------+------------+
| name | owner | species | sex | birth | death |
+----------+--------+---------+------+------------+------------+
| Fluffy | Harold | cat | f | 1993-02-04 | NULL |
| Claws | Gwen | cat | m | 1994-03-17 | NULL |
| Buffy | Harold | dog | f | 1989-05-13 | NULL |
| Fang | Benny | dog | m | 1990-08-27 | NULL |
| Bowser | Diane | dog | m | 1979-08-31 | 1995-07-29 |
| Chirpy | Gwen | bird | f | 1998-09-11 | NULL |
| Whistler | Gwen | bird | NULL | 1997-12-09 | NULL |
| Slim | Benny | snake | m | 1996-04-29 | NULL |
| Puffball | Diane | hamster | f | 1999-03-30 | NULL |
+----------+--------+---------+------+------------+------------+
执行下面的语句:
mysql> SELECT name, birth, CURDATE(),
-> (YEAR(CURDATE())-YEAR(birth))
-> - (RIGHT(CURDATE(),5)<RIGHT(birth,5))
-> AS age
-> FROM pet;结果是:
+----------+------------+------------+------+
| name | birth | CURDATE() | age |
+----------+------------+------------+------+
| Fluffy | 1993-02-04 | 2003-08-19 | 10 |
| Claws | 1994-03-17 | 2003-08-19 | 9 |
| Buffy | 1989-05-13 | 2003-08-19 | 14 |
| Fang | 1990-08-27 | 2003-08-19 | 12 |
| Bowser | 1989-08-31 | 2003-08-19 | 13 |
| Chirpy | 1998-09-11 | 2003-08-19 | 4 |
| Whistler | 1997-12-09 | 2003-08-19 | 5 |
| Slim | 1996-04-29 | 2003-08-19 | 7 |
| Puffball | 1999-03-30 | 2003-08-19 | 4 |
+----------+------------+------------+------+
上面的语句的执行结果是求是宠物的年龄。问题:
第三行我看不懂。第三行是不是多余呢?两个年相减不就是年龄了吗?还有,能帮我解释下第三句的含义吗?谢谢!
+----------+--------+---------+------+------------+------------+
| name | owner | species | sex | birth | death |
+----------+--------+---------+------+------------+------------+
| Fluffy | Harold | cat | f | 1993-02-04 | NULL |
| Claws | Gwen | cat | m | 1994-03-17 | NULL |
| Buffy | Harold | dog | f | 1989-05-13 | NULL |
| Fang | Benny | dog | m | 1990-08-27 | NULL |
| Bowser | Diane | dog | m | 1979-08-31 | 1995-07-29 |
| Chirpy | Gwen | bird | f | 1998-09-11 | NULL |
| Whistler | Gwen | bird | NULL | 1997-12-09 | NULL |
| Slim | Benny | snake | m | 1996-04-29 | NULL |
| Puffball | Diane | hamster | f | 1999-03-30 | NULL |
+----------+--------+---------+------+------------+------------+
执行下面的语句:
mysql> SELECT name, birth, CURDATE(),
-> (YEAR(CURDATE())-YEAR(birth))
-> - (RIGHT(CURDATE(),5)<RIGHT(birth,5))
-> AS age
-> FROM pet;结果是:
+----------+------------+------------+------+
| name | birth | CURDATE() | age |
+----------+------------+------------+------+
| Fluffy | 1993-02-04 | 2003-08-19 | 10 |
| Claws | 1994-03-17 | 2003-08-19 | 9 |
| Buffy | 1989-05-13 | 2003-08-19 | 14 |
| Fang | 1990-08-27 | 2003-08-19 | 12 |
| Bowser | 1989-08-31 | 2003-08-19 | 13 |
| Chirpy | 1998-09-11 | 2003-08-19 | 4 |
| Whistler | 1997-12-09 | 2003-08-19 | 5 |
| Slim | 1996-04-29 | 2003-08-19 | 7 |
| Puffball | 1999-03-30 | 2003-08-19 | 4 |
+----------+------------+------------+------+
上面的语句的执行结果是求是宠物的年龄。问题:
第三行我看不懂。第三行是不是多余呢?两个年相减不就是年龄了吗?还有,能帮我解释下第三句的含义吗?谢谢!
解决方案 »
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