I have an array of links to the site where the screenshot was taken, of this type: http://prnt.sc/c0jkrl

I need to get the link that they have in body stored on the image itself http://image.prntscr.com/image/f81b70e684b64ddf8ef64e1e2334a5d6.png in general, the link can be obtained simply by clicking on "copy the link to the image" of such links I have 800 pieces. Hands do not do well.

How do I programmatically get the best? Sites never parsed, but at the level of the algorithm, everything is clear. You can leave references that look best in my case. Language - Python

Closed due to the fact that it is necessary to reformulate the question so that it was possible to give an objectively correct answer by the participants aleksandr barakin , user194374, lexxl , Vartlok , Streletz Aug 9 '16 at 14:42 .

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1 answer 1

 import requests import lxml.html url = 'http://prnt.sc/c0jkrl' r = requests.get(url) html = lxml.html.fromstring(r.text) img = html.xpath("//*[@name='twitter:image:src']/@content")[0]

Dependencies, lxml, requests

A disadvantage, there is no error handling, if there is no image or the file is not received. Functions did not, the last 3 lines are wrapped in a function and all links are passed by them.