🔒 using link replacements, but doesn't pick up multiple links in one message

[lone fiber]

i'm using the following code that makes the bot reply to the message with the x.com link from the message but replaced with fxtwitter.com, but only picks up the first link sent.

how could i modify this code for the bot to reply with ALL links in the message and send them in a reply with each link on a separate line?

help is much appreciated, thank you very much.

if 'twitter.com' in message.content and 'fxtwitter.com' not in message.content and 'fixupx.com' not in message.content and 'vxtwitter.com' not in message.content:
        urls = re.findall(r'http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\\(\\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+', message.content)        
        await message.edit(suppress=True)
        await message.reply(urls[0].replace("twitter.com", "fxtwitter.com"), mention_author=False) 
        await asyncio.sleep(3)
        await message.edit(suppress=True) #in case the embed doesn't get removed from the original message the first time, this happened many times and this fixes that issue

[void urchin]

@lone fiber

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[proven iris]

Are you familiar with for loops?

Also, how do you handle a case where you have both a twitter.com link and a fxtwitter.com link

[lone fiber]

not really man, ive never really coded. let me look it up. thanks

[lone fiber]

ill cross that bridge when i get to it

blobgrimacing

[lone fiber]

so ive tried

for url in urls:
  await message.reply(url.replace("x.com", "fxtwitter.com"), mention_author=False)```

this sends the links in seperate replies, is there a way to make it all links in one reply, each link on its own line, perhaps?

[proven iris]

yes, instead of sending a reply, append the result of the replace to a list

Then

!join

[void urchin]

Joining iterables

If you want to display a list (or some other iterable), you can write:

colors = ['red', 'green', 'blue', 'yellow']
output = ""
separator = ", "
for color in colors:
    output += color + separator
print(output)
# Prints 'red, green, blue, yellow, '

However, the separator is still added to the last element, and it is relatively slow.

A better solution is to use str.join.

colors = ['red', 'green', 'blue', 'yellow']
separator = ", "
print(separator.join(colors))
# Prints 'red, green, blue, yellow'

An important thing to note is that you can only str.join strings. For a list of ints,
you must convert each element to a string before joining.

integers = [1, 3, 6, 10, 15]
print(", ".join(str(e) for e in integers))
# Prints '1, 3, 6, 10, 15'

[proven iris]

Then send that as a reply

Instead of ", " in the join use "\n" for a newline

[lone fiber]

yeah, this here seems to have done the job

if 'x.com' in message.content and 'fxtwitter.com' not in message.content and 'fixupx.com' not in message.content and 'vxtwitter.com' not in message.content and '/media' not in message.content and '/novx' not in message.content and '/nofx' not in message.content:
        urls = re.findall(r'http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\\(\\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+', message.content)        
        await message.edit(suppress=True)
        test = ""
        for url in urls:
          test += url.replace("x.com", "fxtwitter.com") + "\n"
        await message.reply(test, mention_author=False)
        await asyncio.sleep(3)
        await message.edit(suppress=True)

[proven iris]

nice

[void urchin]

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