How can I get the page source, without showing the page opened, using selenium with chromedriver and python?

Question:

I’m using selenium with Chrome driver; How can I get the page source, without showing the page opened? What I should specify in webdriver.ChromeOptions()?
Here the code:

from selenium.common.exceptions import WebDriverException
from selenium.webdriver.chrome.options import Options

chrome_options = Options()
chrome_options.add_argument("???")
bowser = webdriver.Chrome(chrome_options=chrome_options)


browser = webdriver.Chrome() 
try:
    browser.get("www.google.com")
    html_content = browser.page_source
    #do stuff
    browser.quit()
except WebDriverException:
    print "Invalid URL"
Asked By: d3llafr33

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Source

Answers:

You should not use ChromeDriver but some headless Webdriver like HtmlUnitDriver, explained here

Answered By: Würgspaß

If you are adamant to use selenium, then you can use any of the headless browsers such as htmlunit driver.
Else you can can just send a get request on the URL and get the response text.

Answered By: user3262242

Selenium / Chrome has a headless option, which allows you to load webpages from code:

from selenium.webdriver import Chrome
from selenium.webdriver.chrome.options import Options
chrome_options = Options()
chrome_options.add_argument("--headless")
browser = Chrome(options=chrome_options, executable_path='path_to_chromedriver')
browser.get('https://wwww.mywebsite.com')
Answered By: Peurke