是否有一种解决方法可以防止特定方法将内容打印到控制台？

Question

when I'm scraping details for specific apps in the App Store, it prints out a URL to the console even when there isn't an print statement. Here's my experimental code snippet and the current output.

!pip install itunes-app-scraper-dmi

from itunes_app_scraper.scraper import AppStoreScraper

print(" ")
bundleID = "com.marmalade.monopoly"
appDetails = AppStoreScraper().get_app_details(bundleID, country="us")
version = appDetails['version']
url = appDetails['trackViewUrl'][:-5]

print("\nVersion:", version)
print("App URL:", url)

And it outputs the following.

Looking in indexes: https://pypi.org/simple, https://us-python.pkg.dev/colab-wheels/public/simple/
Requirement already satisfied: itunes-app-scraper-dmi in /usr/local/lib/python3.7/dist-packages (0.9.4)
 
https://itunes.apple.com/lookup?bundleId=com.marmalade.monopoly&country=us&entity=software

Version: 1.7.15
App URL: https://apps.apple.com/us/app/monopoly-classic-board-game/id1477966166

I plan to merge this into my main code which will involve iterating through a larger list of apps, and I'm not interested in the URL that gets printed to the console. I'm only interested in grabbing couple pieces of info for each app and storing them into variables for further processing.

In the code example, it prints this URL to the console with or without independent print statements. ( https://itunes.apple.com/lookup?bundleId=com.marmalade.monopoly&country=us&entity=software )

I looked at the get_app_details() method in (2) below and couldn't find where it prints the URL. In here, I'm looking at Line 177 and anything after, where it puts together the URL based on the given bundle ID. Line 180 seems to be storing the result of requests.get(url).json() into the variable result . Could this sub-method be responsible for printing the URL without its own print statement?

If so, is there a workaround to prevent the URL from being printed to the console?

Documentation

• (1) itunes-app-scraper Documentation
• (2) Methods for itunes-app-scraper

Answer 1

Quick update, I managed to find a solution for this. I used the following code snippets throughout my main code, when I didn't want the method to print the URL.

from IPython.utils import io

with io.capture_output() as captured:
    print("Output Text")