可以使用条件聚合来获得所需的结果。我假设您希望使用每个月的平均评级进行比较。

SELECT u.url, 
       AVG(CASE WHEN MONTH(r.created_at) = MONTH(CURDATE() - INTERVAL 1 MONTH)
                 AND YEAR(r.created_at)  = YEAR(CURDATE() - INTERVAL 1 MONTH) THEN rating END) AS `previous rating`,
       AVG(CASE WHEN MONTH(r.created_at) = MONTH(CURDATE())
                 AND YEAR(r.created_at) = YEAR(CURDATE()) THEN rating END) AS `current rating`
FROM urls u
LEFT JOIN ratings r
ON r.url_id = u.id
GROUP BY u.url
ORDER BY `current rating` DESC

输出：

url         previous rating     current rating
google.com  10.0000             11.0000
bing.com                        6.0000
apple.com   8.0000              5.0000

更新
只使用每个月的最后一次评分（如评论中所要求的）要复杂一些。两个就可以了 JOIN 到包含上个月和本月最后评级的表：

SELECT u.url,
       pr.rating AS `previous rating`,
       cr.rating AS `current rating`
FROM urls u
LEFT JOIN (SELECT url_id, rating
           FROM ratings r1
           WHERE created_at = (SELECT MAX(created_at)
                               FROM ratings r2
                               WHERE r2.url_id = r1.url_id AND
                                   MONTH(created_at) = MONTH(CURDATE() - INTERVAL 1 MONTH) AND
                                   YEAR(created_at)  = YEAR(CURDATE() - INTERVAL 1 MONTH)
                               )
           ) pr ON pr.url_id = u.id
LEFT JOIN (SELECT url_id, rating
           FROM ratings r1
           WHERE created_at = (SELECT MAX(created_at)
                               FROM ratings r2
                               WHERE r2.url_id = r1.url_id
                               )
           ) cr ON cr.url_id = u.id
ORDER BY `current rating` DESC

更新了dbfiddle的演示