我知道了,原来我是在php代码中执行,没得到正确结果,后来在DOS命令提示符下使用就可以了,但现在又发现新问题了.当我按那页面的代码测试后.
SELECT * FROM shirt;
+----+---------+--------+-------+
| id | style | color | owner |
+----+---------+--------+-------+
| 1 | polo | blue | 1 |
| 2 | dress | white | 1 |
| 3 | t-shirt | blue | 1 |
| 4 | dress | orange | 2 |
| 5 | polo | red | 2 |
| 6 | dress | blue | 2 |
| 7 | t-shirt | white | 2 |
+----+---------+--------+-------+,这结果是正确的,然后我用下面的php插入一条记录mysql_query("INSERT INTO shirt VALUES(NULL,'dress','orange',LAST_INSERT_ID())");,在这php执行后,我又转到DOS命令提示符下,在用INSERT INTO shirt VALUES(NULL,'dress','orange',LAST_INSERT_ID());语句执行了一下.结果现在SELECT * FROM shirt;
+----+---------+--------+-------+
| id | style | color | owner |
+----+---------+--------+-------+
| 1 | polo | blue | 1 |
| 2 | dress | white | 1 |
| 3 | t-shirt | blue | 1 |
| 4 | dress | orange | 2 |
| 5 | polo | red | 2 |
| 6 | dress | blue | 2 |
| 7 | t-shirt | white | 2 |
| 8 | dress | orange | 0 |
| 9 | dress | orange | 4 |
+----+---------+--------+-------+这owner的指怎么会为0或4呢.在person表的id中没有这两个值呀?
SELECT * FROM shirt;
+----+---------+--------+-------+
| id | style | color | owner |
+----+---------+--------+-------+
| 1 | polo | blue | 1 |
| 2 | dress | white | 1 |
| 3 | t-shirt | blue | 1 |
| 4 | dress | orange | 2 |
| 5 | polo | red | 2 |
| 6 | dress | blue | 2 |
| 7 | t-shirt | white | 2 |
+----+---------+--------+-------+,这结果是正确的,然后我用下面的php插入一条记录mysql_query("INSERT INTO shirt VALUES(NULL,'dress','orange',LAST_INSERT_ID())");,在这php执行后,我又转到DOS命令提示符下,在用INSERT INTO shirt VALUES(NULL,'dress','orange',LAST_INSERT_ID());语句执行了一下.结果现在SELECT * FROM shirt;
+----+---------+--------+-------+
| id | style | color | owner |
+----+---------+--------+-------+
| 1 | polo | blue | 1 |
| 2 | dress | white | 1 |
| 3 | t-shirt | blue | 1 |
| 4 | dress | orange | 2 |
| 5 | polo | red | 2 |
| 6 | dress | blue | 2 |
| 7 | t-shirt | white | 2 |
| 8 | dress | orange | 0 |
| 9 | dress | orange | 4 |
+----+---------+--------+-------+这owner的指怎么会为0或4呢.在person表的id中没有这两个值呀?
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