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。
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;
在本教程中,我们将使用 web3 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 | 芝麻教程 | http://www.web3.xin/ | 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)
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;执行以上 SQL 输出结果如下:
mysql> 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; +----------+---------------------------+------+ | name | url | nums | +----------+---------------------------+------+ | Facebook | https://www.facebook.com/ | 750 | | Google | https://www.google.cm/ | 275 | | 芝麻教程 | http://www.web3.xin/ | 521 | +----------+---------------------------+------+ 3 rows in set (0.01 sec)
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;执行以上 SQL 输出结果如下:
mysql> select websites.name, websites.url, sum(access_log.count) 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; +----------+---------------------------+-----------------------+ | name | url | sum(access_log.count) | +----------+---------------------------+-----------------------+ | Facebook | https://www.facebook.com/ | 750 | | Google | https://www.google.cm/ | 275 | +----------+---------------------------+-----------------------+ 2 rows in set (0.01 sec)