I am getting the following message : with no data returned Status code = 7. Please help

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نشر بواسطة aniketism_tmdb
STAFFMOD
في أكتوبر 25, 2023 عند الساعة 9:16 مساءا

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

8 ردود (على هذه الصفحة 1 من 1)

رد بواسطة robbie3999

STAFFMOD

بتاريخ أكتوبر 26, 2023 في 12:54 مساءا

Hi @aniketism_tmdb, your missing the string "api_key=" in the url. The url is

https://api.themoviedb.org/3/discover/movie?xxxxxx&primary_release_date.gte=2020-01-01&primary_release_date.lte=2020-02-27

You need to set the string api_key to "api_key=????". Or add "api_key=" to the end of "discover_api".

رد بواسطة aniketism_tmdb

STAFFMOD

بتاريخ أكتوبر 26, 2023 في 2:56 مساءا

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

رد بواسطة superboy97

STAFFMOD

بتاريخ أكتوبر 26, 2023 في 4:22 مساءا

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.

رد بواسطة aniketism_tmdb

STAFFMOD

بتاريخ أكتوبر 26, 2023 في 4:44 مساءا

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......

t🇧

رد بواسطة ticao2 🇧🇷 pt-BR

STAFFMOD

بتاريخ أكتوبر 26, 2023 في 4:45 مساءا

A suggestion to try to resolve it.
Try changing your line below

discover_api = "https://api.themoviedb.org/3/discover/movie?"

to this line below

discover_api = "https://api.themoviedb.org/3/discover/movie?api_key="

رد بواسطة robbie3999

STAFFMOD

بتاريخ أكتوبر 26, 2023 في 4:46 مساءا

Yes, you are missing the point, your code produces this url:

https://api.themoviedb.org/3/discover/movie?xxxxxx&primary_release_date.gte=2020-01-01&primary_release_date.lte=2020-02-27

You need it to produce this url:

https://api.themoviedb.org/3/discover/movie?api_key=xxxxxx&primary_release_date.gte=2020-01-01&primary_release_date.lte=2020-02-27

You need to add the string "api_key=" to one of the variables or the statement that creates the url.

رد بواسطة aniketism_tmdb

STAFFMOD

بتاريخ أكتوبر 26, 2023 في 4:50 مساءا

@ticao2 said:

A suggestion to try to resolve it.
Try changing your line below
discover_api = "https://api.themoviedb.org/3/discover/movie?"
to this line below
discover_api = "https://api.themoviedb.org/3/discover/movie?api_key="

Thanks robbie

رد بواسطة aniketism_tmdb

STAFFMOD

بتاريخ أكتوبر 26, 2023 في 4:50 مساءا

@ticao2 said:

A suggestion to try to resolve it.
Try changing your line below
discover_api = "https://api.themoviedb.org/3/discover/movie?"
to this line below
discover_api = "https://api.themoviedb.org/3/discover/movie?api_key="

Thanks ticao2

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