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Replying with a image or gif obtained via API, in telegram bot python (telegram.ext)

I am building a meme generator telegram bot that sends a meme whenever the user commands /meme. I am using an API, say "https://......meme-api.....". It gives me a URL for the meme. I want to send the media in that URL provided by the API, as a reply to the /meme command.
dispatcher.add_handler(telegram.ext.CommandHandler("meme", meme))
def meme(update, context):
response = requests.get('https://meme-api.herokuapp.com/gimme').json()
url = response.get('url')
print(url)
update.message.reply_text(url)
I can't wrap my head around how I can write the meme function. What message reply type should I use to send the media without saving it?

As explained in the documentation of Message.reply_text that method is a shortcut for Bot.send_message, which corresponds to sendMessage in the Telegram docs. This method sends text messages.
To send photos & animantions, you'll have to use
Message.reply_photo / Bot.send_photo / sendPhoto
Message.reply_animantion / Bot.send_animantion / sendAnimantion,
respectively.

Is there a way to access every website's data through discord.py

Hi stalkers
Is there a way to access a site's data directly?
I need it for my code :
#commands.command(aliases = ['isitsafe','issafe','scanlink'])
async def isthissafe(self, ctx, link: str):
try:
link = 'https://transparencyreport.google.com/safe-browsing/search?url='+ link.replace('/','%2F')
embed=discord.Embed(
color = discord.Color.dark_red(),
title = '',
description = f"[Transparency Report verification]({link})")
await self.emb(embed, ctx.author.name, 'https://cwatch.comodo.com/images-new/check-my-site-security.png')
await ctx.send(embed=embed)
except:
await ctx.send('An error has occured')
print('\nERROR')
Basically I made, a command which should tell if a link is safe or not, I did it using google's verification report site, but.. the problem is I only reformatted the link so the bot sens it in an embed and you access it from there.
My question is, now that you understood what I need, is there some way in which I could directly let the bot output the message from the website that indicates if the site is malicious/safe ??
Please help me.
I provided an image as well with the message I want to get from the site.

You might want to try scraping the site with bs4, or just look for the string "No unsafe content found". However, it looks like google populates the field based on a request.
Your best bet would be to use transparencyreport.google.com/transparencyreport/api/v3/safebrowsing/status?site=SITE_HERE. It returns a JSON response, but I don't understand it, so play around and figure out what the keys mean

create email with python

im working in a personal project that needs new email in the start, and i want create a new email with python also i don't want run a complicate smtp server(I don't know much about that yet) i want do something like temp mail with api, i'd tried temp mail api but i got error i do something like this
import requests
url = "privatix-temp-mail-v1.p.rapidapi.com/request/mail/id/md5 of my temp mail"
req = request.get(url)
print(req)
but i got 401 status code that says your api key is invalid
then i go to rapidapi website and see examples there was a header for req so i put that to my code that was like:
import requests
url = "https://privatix-temp-mail-v1.p.rapidapi.com/request/mail/id/md5"
headers = {
'x-rapidapi-host': "privatix-temp-mail-v1.p.rapidapi.com",
'x-rapidapi-key': "that was a key"
}
req = request.get(url, headers=headers)
then i got this
{"message":"You are not subscribed to this API."}
now i get confused and i don't know what is problem if you know temp mail api or something liks this service or any suggest pls help me

In order to use any API from RapidAPI Hub, you need to subscribe to that particular API. It's pretty simple.
Go to the Pricing Page of this API and choose a plan according to your need. Click on the subscribe button and you will be good to go. However, the Basic plan is free but a soft limit is associated with it so it may ask for your card details.

How can I use the output of a python script as RASA’s input instead of the usual user input?

I am currently using RASA and developed a working chatbot. One part of my project is to use a speech-to-text recognition, and I wrote a working code in Python that returns the text said by the user.
I want to use that text for RASA’s input, instead of writing like usual.
I saw there was something to do with the inputs channels, but I only saw input that are other webservices and couldn’t figure it out for using just a local script.
Thank you for any advice,
LM

You can try rasa REST API for this purpose. Make sure that you have action_endpoint url in endpoints.yml. Normally it is
url: "http://localhost:5055/webhook"
Then make sure your rasa bot is up and if htere are any custom actions, start that server as well.
After starting your webhook, you can simple call
http://localhost:5005/webhooks/rest/webhook
and in the payload you have to put below payload
messagePayload = {
sender: 'default',
message: 'Your message is here'
}
and finally add httpheader content type as application/json like below
'Content-Type': 'application/json'
Now your bot will work fine.
tldr;
If you are using request in your python for api calls, you can try below code.
import requests
API_ENDPOINT = "http://localhost:5005/webhooks/rest/webhook"
messagePayload = {
sender: 'default',
message: 'Your message is here'
}
r = requests.post(url = API_ENDPOINT, data = messagePayload)

How about just using Rest API that is already present in the library.
For this, you just need to fill the query parameter, which you can do with your script, rather than writing a custom Input Channel.

Using Slack RTM API with Django 2.0

I am a beginner to the Django framework and I am building a Django app that uses the Slack RTM API.
I have a coded a program in python that performs the OAuth authentication process like so :
def initialize():
url="https://slack.com/api/rtm.connect"
payload={"token":"xxx"}
r=requests.post(url,payload)
res=json.loads(r.text)
url1=res['url']
ws = create_connection(url1)
return ws
My Requirement:
The stream of events I receive (from my slack channel that my slack app is added to) is processed to filter out events of the type - message ,then match the message with a regex pattern and then store the matched string in a database.
As a stand alone python program I am receiving the stream of events from my channel.
My questions:
How do I successfully integrate this code to Django so that I can
fulfill my requirement?
Do I put the code in templates/views? What is the
recommended method to process this stream of data?

def initialize():
url = "https://slack.com/api/rtm.connect"
r = requests.get(url, params={'token': '<YOUR TOKEN>'})
res = r.json()
url1=res['url']
ws = create_connection(url1) #Note: There is no function called create_connnection() so it will raise an error
return ws
if you read the API web methods, you see :
Preferred HTTP method: GET
See here: Slack rtm.connect method
look at the comment, and thats the right code, see the differences between this code and yours.
basically to get JSON from a request don't use json.loads because this search your local computer not the request
use r.json() so it call the json you got from r.
Note that r.text will return raw text output so when you want to get url it will not be identified, with r.json you can call the object url as stated about
Hope this help.
and please could you tell us more what you wanna do with this in view ? because template is a directory which contains all the HTML files which you don't need to work with.
but why views.py ?