To store multiple URLs efficiently, save them in a text file and read each line one by one. Then capture a screenshot using the host name from the URL as part of the file name.
I have made adjustments to your code so that each URL's screenshot is saved separately. My implementation was done using Python 3.6.9.
Here is how my directory structure looks:
.
├── links.txt
├── requirements.txt
└── screenshots_of_links.py
The content of links.txt is as follows:
http://facebook.com
http://apple.com
The contents of requirements.txt are:
selenium==3.141.0
urllib3==1.25.10
The contents of screenshots_of_links.py script are shown below:
from selenium import webdriver
from urllib.parse import urlparse
from time import sleep
driver = webdriver.Firefox()
with open("links.txt") as url_file:
for line in url_file.readlines():
url = line.strip()
if url != "":
driver.get(url)
host = urlparse(url).hostname
driver.get_screenshot_as_file("{}.png".format(host))
sleep(2)
driver.quit()
print("All tasks completed successfully...")
Generated Output:
https://i.stack.imgur.com/QAmRt.png
https://i.stack.imgur.com/uwFiS.png
Details of Modifications Made:
- Read the URLs stored in
links.txt.
- Trimmed any extra spaces from each line.
- Used the hostname extracted from each URL as the filename for corresponding screenshot captures. The function
urlparse(url).hostname was used for this purpose.
Further Reading: