Issue
I have a URL and when I send a request by curl, I get a big output.
curl https://www.aparat.com/video/video/embed/videohash/lXhkG/vt/frame -H "Accept: application/json" -s
I get: https://pastebin.mozilla.org/QM6FN8MZ#L
But I just want to get the URL of 720p, I mean just:
https:\/\/caspian1.cdn.asset.aparat.com\/aparat-video\/de54245e862b62249b6b7958c734276547445778-720p.apt?wmsAuthSign=eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJ0b2tlbiI6IjQ2NDJhYmQ4NGFiN2UzNDJkNGMxZWI3ZTNkMzlmZmQ5IiwiZXhwIjoxNjY5ODA5NzI1LCJpc3MiOiJTYWJhIElkZWEgR1NJRyJ9.havkkhJyXjBt_jHPVv4poEVb65_7tRsLIxO5pCO7tGE
Any idea how to do it?
I'm trying to use grep but I don't know how to remove other things from else 720p URL.
curl https://www.aparat.com/video/video/embed/videohash/lXhkG/vt/frame -H "Accept: application/json" -s | grep -e "720p"
Solution
I would harness GNU AWK for this following way
wget --quiet -O - https://www.aparat.com/video/video/embed/videohash/lXhkG/vt/frame | awk 'match($0, /http[^"]*720[^"]*/){print substr($0,RSTART,RLENGTH)}'
gives output
https:\/\/caspian1.cdn.asset.aparat.com\/aparat-video\/de54245e862b62249b6b7958c734276547445778-720p.apt?wmsAuthSign=eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJ0b2tlbiI6IjY1OTcxYTRkNGZiMjkyYjk0NjM0Mjk2ODVkOTc3YjEwIiwiZXhwIjoxNjY5ODIxNDM2LCJpc3MiOiJTYWJhIElkZWEgR1NJRyJ9.NI2_6nwOxLEOxhWghsR2bOqzrXINXqqscbduHpCWwok
Explanation: I use wget with information like progress bar &c turned-off (--quiet) and writing to standard output (-O -) which is piped into awk, which for each line is matching against following regular expression http[^"]*720[^"]* that is http followed by zero-or-more (*) not-quotes followed by 720 followed by zero-or-more non-quotes, if there is match I print substring of line containing that match. match string function sets RSTART and RLENGTH variables, which I use later in substr. Note: this might give false positivie if there are others URL containing 720.
(tested in GNU Wget 1.20.3 and GNU Awk 5.0.1)
Answered By - Daweo Answer Checked By - Mildred Charles (WPSolving Admin)
