SQL HAVING 子句

HAVING 子句

在 SQL 中增加 HAVING 子句原因是，WHERE 关键字无法与聚合函数一起使用。

HAVING 子句可以让我们筛选分组后的各组数据。

SQL HAVING 语法

SELECT column_name, aggregate_function(column_name)
FROM table_name
WHERE column_name operator value
GROUP BY column_name
HAVING aggregate_function(column_name) operator value;

演示数据库

在本教程中，我们将使用 RUNOOB 样本数据库。

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  | 教程      | http://www.itpon.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 |
+-----+---------+-------+------------+
9 rows in set (0.00 sec)

SQL HAVING 实例

现在我们想要查找总访问量大于 200 的网站。

我们使用下面的 SQL 语句：

SELECT Websites.name, Websites.url, SUM(access_log.count) AS nums FROM (access_log
INNER JOIN Websites
ON access_log.site_id=Websites.id)
GROUP BY Websites.name
HAVING SUM(access_log.count) > 200;

现在我们想要查找总访问量大于 200 的网站，并且 alexa 排名小于 200。

我们在 SQL 语句中增加一个普通的 WHERE 子句：

SELECT Websites.name, SUM(access_log.count) AS nums FROM Websites
INNER JOIN access_log
ON Websites.id=access_log.site_id
WHERE Websites.alexa < 200 
GROUP BY Websites.name
HAVING SUM(access_log.count) > 200;