Content Curation Service
Scoop.it is a content curation service that helps professionals and businesses research and publish content.
import requests
api_key = 'YOUR_API_KEY'
# Note: The actual API endpoint for Scoop.it is not provided in the description.
# This is a hypothetical example for demonstration.
url = f'https://api.scoop.it/v1/topics?api_key={api_key}'
response = requests.get(url)
response.raise_for_status() # Raise an exception for HTTP errors
print(response.json())Last 50 checks (10 min intervals)