Telegram bot should send URL that contains "&" character, but it sends the URL only until that character and then cuts it short

Question

I want my Telegram bot to send a URL to a channel. However, the url contains the "&" character which cuts the message that it's trying to send short. The Telegram API documentation says I need to use & amp; (without the space) to replace & but either I don't understand something or it doesn't work.

Here's what I'm doing:

requests.get("https://api.telegram.org/"+botID+"/sendMessage?chat_id="+chatid+"&text="+movieSearch+"&parse_mode=HTML")

And movieSearch is:

movieSearch = ("https://www.imdb.com/search/title?release_date="+year+"-01-01,2018-12-31&user_rating="+score+",&genres="+genres)

You can see that in movieSearch after the release_date there is &user_rating=... and so on. However, the bot will only send the URL until just before that & character (so until "2018-12-31").

I've tried replacing & with & amp; but it still won't send the whole URL. I've tried without the parse_mode=HTML but it didn't work either.

@johnashu Here's an example of a real URL and it has the comma too, I tried removing it but it doesn't have an effect: https://www.imdb.com/search/title?release_date=1990-01-01,2018-12-31&user_rating=8.0,&genres=action,adventure,comedy — KMFR, May 10 '18 at 21:24
What happens when you try the link you posted instead of `movieSearch` into the request.. — johnashu, May 10 '18 at 21:30
@johnashu it sends: https://www.imdb.com/search/title?release_date=1990-01-01,2018-12-31 — KMFR, May 10 '18 at 21:33
@PauloScardine if I simply replace & with `&`, it does the same as with & — KMFR, May 10 '18 at 21:34
You must escape it as `%26amp;` instead. This is because you are not letting the requests library escape it for you. — Paulo Scardine, May 10 '18 at 21:36
@PauloScardine wow, `%26amp;` worked. Had no idea the requests library had an effect, thank you! — KMFR, May 10 '18 at 21:43

Paulo Scardine · Accepted Answer · 2018-05-11T19:27:48.987

Instead of:

requests.get("https://api.telegram.org/"+botID+"/sendMessage?chat_id="
             +chatid+"&text="+movieSearch+"&parse_mode=HTML")

Do this:

params = {
    "chat_id": chatid,
    "text": movieSearch,
    "parse_mode": "HTML",
}

requests.get(
    "https://api.telegram.org/{}/sendMessage".format(botID),
    params=params
)

I think the problem happens because you have & in the value of the "text" parameter in the URL but are not escaping it as %26. It is better to use a dictionary instead and let the requests library escape it for you. You still have to escape the & as &:

movieSearch = "https://www.imdb.com/search/title?release_date={}"
              "-01-01,2018-12-31&amp;user_rating={}&amp;genres={}".format(
                    year, score, genres)

Alright thank you sir. Escaping as `%26amp;` works, I'm gonna use the dictionary from now on as well. — KMFR, May 10 '18 at 21:47

score 0 · Answer 2 · answered Mar 06 '21 at 09:49

0

you sholud use the ASCII Encoding of char.

more info: URL Encoding

answered Mar 06 '21 at 09:49

AminRostami

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Telegram bot should send URL that contains "&" character, but it sends the URL only until that character and then cuts it short

2 Answers2