urllib bad statusline in python

To handle the "BadStatusLine" error with urllib in Python, you can use a try-except block when sending the HTTP request. Here's an example:

main.py
import urllib.request
import http.client

try:
    # send the HTTP GET request using urllib
    response = urllib.request.urlopen('http://example.com')
    print(response.read())  # print the response content
except http.client.BadStatusLine as e:
    print("BadStatusLine error: ", e)
    # handle the error as needed
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In the code above, we are sending an HTTP GET request to "http://example.com" using urllib.request.urlopen(). If we encounter a "BadStatusLine" error (which occurs when the server sends an invalid HTTP response), we catch it using an except block and print an error message. You can replace the print() statement with your custom error-handling code.