I am using the correct API key. And when I use that api key with a sample code in web browser, it fetches the correct result but with the code (below) and my query, it does not
Here is the code :
import pandas as pd
import requests
pd.options.display.max_columns = 30
discover_api = "https://api.themoviedb.org/3/discover/movie?"
api_key="????"
query1 = "&primary_release_date.gte=2020-01-01&primary_release_date.lte=2020-02-27"
url = discover_api+api_key+query1
data = requests.get(url).json()
data
Output :
{'status_code': 7,
'status_message': 'Invalid API key: You must be granted a valid key.',
'success': False}
Can somebody help me, to understand, what is wrong with the code, as its not fetching any data
Thanks for your help in advance
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robbie3999 的回复
于 2023 年 10 月 26 日 12:54下午
Hi @aniketism_tmdb, your missing the string "api_key=" in the url. The url is
You need to set the string api_key to "api_key=????". Or add "api_key=" to the end of "discover_api".
aniketism_tmdb 的回复
于 2023 年 10 月 26 日 2:56下午
Hi Robbie,
Thanks for responding. Yes I did set the variable api_Key to the real api key, instead of ????
Also, instead I did try to add it to the end of discover_api as discover_api = "https://api.themoviedb.org/3/discover/movie?" But still getting the same message as 'Invalid API key: You must be granted a valid key.
Any other information will help
Thanks
superboy97 的回复
于 2023 年 10 月 26 日 4:22下午
Is the content of your api_key variable in the form "api_key=YOUR_KEY" ? If the variable just contain the key, this will not worl.
aniketism_tmdb 的回复
于 2023 年 10 月 26 日 4:44下午
Thanks superboy97,
Yes, I couldn't understand that earlier and after the message from robbie and you, I realized that I was missing the "api_key=" string in the variable. Its working now....
Thanks very much... Appreciate everyone,s help
Happy Learning......
ticao2 🇧🇷 pt-BR 的回复
于 2023 年 10 月 26 日 4:45下午
A suggestion to try to resolve it.
Try changing your line below
to this line below
robbie3999 的回复
于 2023 年 10 月 26 日 4:46下午
Yes, you are missing the point, your code produces this url:
You need it to produce this url:
You need to add the string "api_key=" to one of the variables or the statement that creates the url.
aniketism_tmdb 的回复
于 2023 年 10 月 26 日 4:50下午
Thanks robbie
aniketism_tmdb 的回复
于 2023 年 10 月 26 日 4:50下午
Thanks ticao2