这可能是一个愚蠢的问题,不是一个完全与编程相关的问题,但在我尝试学习时请耐心等待。
如何FEEDLY只需在搜索框中输入网站 URL 即可从网站检索提要/新闻。
例如,我输入了http://www.inspirationfeed.com进入搜索字段,传递的 AJAX 调用/查询字符串是
http://cloud.feedly.com/v3/search/feeds?q=http%3A%2F%2Finspirationfeed.com%2F&n=20&ck=1381664838936
它返回一个 JSON 作为响应
{"results":[{"deliciousTags":["Design","inspiration","Diseño","Web Design","webdesign","graphic design"],"lastUpdated":1381561980000,"score":27531.265625,"website":"http://inspirationfeed.com","title":"inspirationfeed.com","subscribers":1772,"language":"en","velocity":16.3,"feedId":"feed/http://feeds.feedburner.com/inspirationfeed/BTLD","description":"be inspired!"},{"deliciousTags":["DESIGN","Inspiration"],"lastUpdated":1381474380000,"score":48.0,"website":"http://inspirationfeed.com","title":"inspirationfeed.com » Inspiration","subscribers":48,"language":"en","velocity":4.7,"feedId":"feed/http://inspirationfeed.com/category/inspiration/feed/","description":"be inspired!"},{"deliciousTags":["design"],"lastUpdated":1381479900000,"score":294.7439270019531,"website":"http://inspirationfeed.com","title":"inspirationfeed.com » Articles","subscribers":27,"language":"en","velocity":11.7,"feedId":"feed/http://inspirationfeed.com/category/articles/feed/","description":"be inspired!"},{"deliciousTags":["Photography"],"lastUpdated":1381129620000,"score":17.0,"website":"http://inspirationfeed.com","title":"inspirationfeed.com » Photography","subscribers":17,"language":"en","velocity":0.7,"feedId":"feed/http://inspirationfeed.com/category/photography/feed/","description":"be inspired!"},{"deliciousTags":["Little Bit of Everything-Amatuer"],"lastUpdated":1381396020000,"score":109.16442108154297,"website":"http://inspirationfeed.com","title":"inspirationfeed.com » Blogging","subscribers":10,"language":"en","velocity":0.5,"feedId":"feed/http://inspirationfeed.com/category/articles/blogging/feed/","description":"be inspired!"}]}
我的理解
1)根据关键字 deliciousTags
,feedly 生成 # 标签
2)根据关键的score
,feedly以自上而下的顺序排列站点可用的不同提要/新闻URL
我不明白的地方
1)feedly 如何通过输入网站 URL 获取新闻提要/rss URL (EXAMPLE : "feedId":"feed/http://feeds.feedburner.com/inspirationfeed/BTLD"
)
2)这种东西需要写什么类型的web服务。是否需要服务器端脚本来深入搜索有问题的网站并检索提要 URLS?可以使用 JQuery/Javascript 在客户端本身获取提要 URL 吗?
再次感谢您花时间阅读问题。
最佳答案
在页面 html 的头部有一个带有 rss/atom feed 的站点的链接,如下所示:
<link rel="alternate" type="application/atom+xml" title="Feed for question 'how does a service like feedly obtain the rss feeds from a website when the website URL is entered into the search box(not the feed url'" href="/feeds/question/19345075">
这是这个问题页面的来源
Feedly 或类似的服务在该页面(而非主机)的 html 中搜索并找到此链接标记(带有 rel,type):<link rel="alternate" type="application/atom+xml" title="awd" href="/feedUrl">
或 <link rel="alternate" type="application/rss+xml" title="awd" href="/feedUrl">
(还有更多)
还有一些设置 feed url 的例程,
-
http://<url>/rss
-
http://<url>/feed
-
http://<url>/atom
-
http://<url>/<page>.xml
如果没有找到,那么那里什么也没有!
它可以在服务器端和客户端完成,但在服务器端有更多的好处,例如在解析页面 html 和测试例程之前进行数据库搜索。
默认情况下,Firefox 有一个获取页面提要的选项,以前版本的 chrome 有这样的功能,但它被谷歌(和谷歌方式)删除了。 (这是浏览器端,由用户控制和拥有,网站不能使用!但您可以找到他们实现该方法的方式)
关于javascript - 当网站 URL 输入搜索框(而不是提要 url)时,像 feedly 这样的服务如何从网站获取 rss 提要,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19345075/