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:
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:
示例
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;