CODEWITHSUNDEEP
pythonpython-requestsgiphygiphy-api
Vihaan Patil
Vihaan Patil

Reputation: 17

I'm using the Giphy API and I was wondering how you could get a GIF's Image URL automatically, not manually

Here is the code I have, it works fine, but how could I make it that there is a new variable that is equal to the Image URL of the GIF so that the user can get the Source URL of the GIF?

import requests

url = "https://giphy.p.rapidapi.com/v1/gifs/search"

searchtag = input()

querystring = {"api_key":"secret_key","q":searchtag,"limit":"1","offset":"0","rating":"pg-13"}

headers = {
    'x-rapidapi-key': "4005b98f9bmsh977c629b89034a7p19b52fjsn22b7fb5ce3bb",
    'x-rapidapi-host': "giphy.p.rapidapi.com"
    }

response = requests.request("GET", url, headers=headers, params=querystring)

print(response.text)

Upvotes: 0

Views: 1350

Answers (1)

tear
tear

Reputation: 11

First off you should probably remove your api key for your own personal safety . I'd recommend looking into environmental files for storing private information that doesn't belong in a database. Second off to answer your question you could just declare the url as a variable if you're feeling ever so inclined, here:

    import requests

url = "https://giphy.p.rapidapi.com/v1/gifs/search"

searchtag = input()

querystring = {"api_key":process.env.APIKEY,"q":searchtag,"limit":"1","offset":"0","rating":"pg-13"}

headers = {
    'x-rapidapi-key': "4005b98f9bmsh977c629b89034a7p19b52fjsn22b7fb5ce3bb",
    'x-rapidapi-host': "giphy.p.rapidapi.com"
    }

response = requests.request("GET", url, headers=headers, params=querystring)
gifurl = response.text
print(gifurl)

Upvotes: 0

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