PROBLEM SET 7: WATCH ON YOUTUBE | SOLUTION (CS50 PYTHON)

[@DorsCodingSchool]

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[@eugenekernan7621]

You have placed an asterisk after the "s" in the pattern for the URL protocol. This would match 0 or more "s" s after "http". Should you not put a "?" after "s" in order to indicate either no "s" or 1 "s" ?

[@DorsCodingSchool]

You're correct. Thank you!

[@andrejg3086]

Good video. You don't need to use a backslash "\" to escape a slash "/", because a slash is not a special character in regular expressions in Python.

[@francisvincejaca4077]

i could've never figured out the pattern without this video. Thank you!

[@DorsCodingSchool]

Glad we could help!

[@federicoelevazo3622]

I got terrible migraine trying to figure out the regex solution for this. If I opt using regex, the solution is easy. But since I wanna learn more with regex, I surrender and end up watching this vid to learn the proper way of doing it. ^_^

[@getstart98]

Splendid as always!

[@athbuys3588]

thanks so much! the way you go through the regex patterns makes the very easy to follow

[@DorsCodingSchool]

We are glad to hear that :)

[@rohith6255]

do you really have to check if the url is inside ?? can't we just match it https... by using caret ^ operator ??

[@PurpleDovee]

no because ^ will mean nothing can be before , but in html files cant be the only tag can it

[@luklach]

you can do def parse in 2 lines instead of 5 :)

[@DorsCodingSchool]

Great 😊

[@ertikrajkova2748]

how

[@IamMuslimChad]

Man i was stuuuck for many hours........ I was just doing from the whole

[@awesomeadvices5962]

super awesome

[@manoellopes]

Salvou ! 👍

[@zhrrbh2048]

r'' it must be r'' bc Adding ? immediately after either, a la *? or +?, “makes it perform the match in non-greedy or minimal fashion; as few characters as possible will be matched.”

[@DorsCodingSchool]

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[@justynyeo4521]

found the easiest way to do it that aligns more with the lecture: import re import sys def main(): print(parse(input("HTML: "))) def parse(s): if match := re.search(r'src="(?:https?://(?:www\.)?youtube\.com/embed/([a-zA-Z0-9]+))"', s): return(f"youtu.be/{match.group(1)}") else: return None if __name__ == "__main__": main()

[@rideaux8533]

i found the most idiotic way to solve this problem and it worked i think i could do anything to avoid using regex def main(): print(parse(input("HTML: "))) def parse(s): try: line = s.find('src="') line = line + 5 qpos = s.find('"', line) final = s[line:qpos] rev1 = final.replace('embed/','') rev2 = rev1.replace('be','') rev3 = rev2.replace('.com','.be') rev4 = rev3.replace('www.','') rev5 = rev4.replace('http','https') rev6 = rev5.replace('httpss','https') if not 'yout' in rev6: return None elif not 'https' in rev6: return None return rev6 except: return None if __name__ == "__main__": main()

[@rutesh1902]

bro ☠☠☠

[@zikobht6618]

Bro has beef with Regex

[@haseifan]

Thanks for this, for some reason the stupid bot fails my solution... gives me: expected output: kzbin.info/www/bejne/rqepi52larWafZI actual output: kzbin.info/www/bejne/rqepi52larWafZI... its broken😅

[@RayhaanKhan-mu4qu]

doesnt everyone?@@zikobht6618

[@RayhaanKhan-mu4qu]

great!

[@PurpleDovee]

import re import sys def main(): print(parse(input("HTML: "))) def parse(s): match = re.search(r'src="(?:https?://(?:www\.)?)?youtube\.com/embed/(?P[^"]+)"', s) if match: vid_id = match.group("vid_id") return f"youtu.be/{vid_id}" return None if __name__ == "__main__": main()

[@muhammadrayyan147]

This is my code, I went with this approach. It works perfectly but could you please tell me if this is the write way or are there any bugs in this import re import sys def main(): print(parse(input("HTML: "))) def parse(s): if gained := re.search(r"^.+(kzbin.info)(\w+)\".+$", s): YT_ID=gained.group(1) + gained.group(2) YT_ID=YT_ID.replace("www.","").replace("be.com/embed",".be") return YT_ID if __name__ == "__main__": main()