下面是选自 "Websites" 表的数据:

+----+--------------+---------------------------+-------+---------+
| id | name         | url                       | alexa | country |
+----+--------------+---------------------------+-------+---------+
| 1  | Google       | https://www.google.cm/    | 1     | USA     |
| 2  | 淘宝          | https://www.taobao.com/   | 13    | CN      |
| 3  | 菜鸟教程      | http://www.runoob.com/    | 4689  | CN      |
| 4  | 微博          | http://weibo.com/         | 20    | CN      |
| 5  | Facebook     | https://www.facebook.com/ | 3     | USA     |
| 7  | stackoverflow | http://stackoverflow.com/ |   0 | IND     |
+----+---------------+---------------------------+-------+---------+

下面是 "access_log" 网站访问记录表的数据:

mysql> SELECT * FROM access_log;
+-----+---------+-------+------------+
| aid | site_id | count | date       |
+-----+---------+-------+------------+
|   1 |       1 |    45 | 2016-05-10 |
|   2 |       3 |   100 | 2016-05-13 |
|   3 |       1 |   230 | 2016-05-14 |
|   4 |       2 |    10 | 2016-05-14 |
|   5 |       5 |   205 | 2016-05-14 |
|   6 |       4 |    13 | 2016-05-15 |
|   7 |       3 |   220 | 2016-05-15 |
|   8 |       5 |   545 | 2016-05-16 |
|   9 |       3 |   201 | 2016-05-17 |
+-----+---------+-------+------------+


表的别名实例

下面的 SQL 语句选取 "菜鸟教程" 的所访问记录。我们使用 "Websites" 和 "access_log" 表,并分别为它们指定表别名 "w" 和 "a"(通过使用别名让 SQL 更简短):

实例

SELECT w.name, w.url, a.count, a.date 
FROM Websites AS w, access_log AS a 
WHERE a.site_id=w.id and w.name="菜鸟教程";

执行输出结果:

不带别名的相同的 SQL 语句:

实例

SELECT Websites.name, Websites.url, access_log.count, access_log.date 
FROM Websites, access_log 
WHERE Websites.id=access_log.site_id and Websites.name="菜鸟教程";

执行输出结果:

posted on 2017-02-28 16:28  芦苇娓娓  阅读(551)  评论(0编辑  收藏  举报