Loop for print screen

Question

Need some help. I got this python code to load url and take a screen print.

I need to achieve this:

1. instead of array, read url from text file
2. take screen print of each url that is loaded and save it. with the code the screen print is over written

from selenium import webdriver
from time import sleep

driver = webdriver.Firefox()
url = ["http://google.com", "http://cisco.com"]
for element in url:
    driver.get(element)
    driver.get_screenshot_as_file("screenshot.png")
sleep(2)
driver.quit()
print("end...")

Answer 1

Store the URLs in a text file and then read line by line. Then take screenshot using file name with the host name of the URL.

I have modified your code and could store the screenshot of each url in separate file. I have used Python 3.6.9.

Directory structure:

.
├── links.txt
├── requirements.txt
└── screenshots_of_links.py

links.txt:

http://google.com
http://cisco.com

requirements.txt:

selenium==3.141.0
urllib3==1.25.10

screenshots_of_links.py:

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("end...")

Output:

Modification details:

• Read the URLs from links.txt text file.
• Trim each line of the file.
• Parsed each URL and used the hostname as filename of screenshot. urlparse(url).hostname returns the hostname of a valid URL.

Reference:

• URL parsing Python documentation