SQL AVG()
AVG() 函数
AVG() 函数返回数值列的平均值。
SQL AVG() 语法
SELECT AVG(column_name) FROM table_name
演示数据库
在本教程中,我们将使用 php 样本数据库。
下面是选自 "access_log" 表的数据:
+-----+---------+-------+------------+
| 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 |
+-----+---------+-------+------------+
| 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 |
+-----+---------+-------+------------+
SQL AVG() 实例
下面的 SQL 语句从 "access_log" 表的 "count" 列获取平均值:
实例
SELECT AVG(count) AS CountAverage FROM access_log;
执行以上 SQL 输出结果如下:
下面的 SQL 语句选择访问量高于平均访问量的 "site_id" 和 "count":
实例
SELECT site_id, count FROM access_log
WHERE count > (SELECT AVG(count) FROM access_log);
WHERE count > (SELECT AVG(count) FROM access_log);
执行以上 SQL 输出结果如下: