Having

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;

访问量大于两百。

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;

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

havaing 对 聚合函数的值 做逻辑判断 》=《