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If html content is loaded using ajax, you have to give the URL of the Ajax request html. See http://apify.heroku.com/resources/4fc2a234ae684d0001000008/edit htt
by sathish316 14y ago
If html content is loaded using ajax, you have to give the URL of the Ajax request html. See http://apify.heroku.com/resources/4fc2a234ae684d0001000008/edit http://apify.heroku.com/resources/4fc2a234ae684d0001000008/e... for example.
It won't work for #! URLs. Twitter has a nice streaming API for search if that's what you're looking for.
- gildas 14y ago* Ajax responses are often JSON content. How do I apply a CSS/XPath selector on it? * I guess #! URLs could be transformed into _escaped_fragment_ URLs [1] * I know twitter has an API. It was just an example. Maybe this example [2] would be more relevant (content could also be fetched with an _escaped_fragment_ URL). [1] https://developers.google.com/webmasters/ajax-crawling/docs/specification https://developers.google.com/webmasters/ajax-crawling/docs/... [2] http://apify.heroku.com/resources/4fcb2f7ba06a160001000044 http://apify.heroku.com/resources/4fcb2f7ba06a160001000044