Is there a way to capture a screenshot of a website based on specified screen size, pixels, or by performing scroll up and down operations? I need to do this similar to using the snipping tool. Can anyone provide me with a solution to achieve this using the following URL?
This is the code snippet I am currently using in AWS Lambda:
import json
#coding=utf-8
import time
from selenium.webdriver import Chrome
from selenium.webdriver.chrome.options import Options
import logging
import os
class WebDriver(object):
def __init__(self):
self.options = Options()
self.options.binary_location = '/opt/headless-chromium'
#self.options.add_argument('--headless')
self.options.add_argument('--no-sandbox')
#self.options.add_argument('--start-maximized')
#self.options.add_argument('--start-fullscreen')
self.options.add_argument('--single-process')
self.options.add_argument('--disable-dev-shm-usage')
self.options.add_argument("--kiosk")
def get(self):
driver = Chrome('/opt/chromedriver', options=self.options)
return driver
def lambda_handler(event, context):
instance_ = WebDriver()
driver = instance_.get()
options = Options()
a = os.listdir('/tmp')
for x in a:
print(x)
URL = os.environ.get("URL")
driver.get(URL)
S = lambda X: driver.execute_script('return document.body.parentNode.scroll'+X)
#driver.set_window_size(S('Width'),S('Height'))
#driver.set_window_size(1280, 720)
# May need manual adjustment
driver.find_element_by_tag_name('body').screenshot('/tmp/job_status.png')
driver.quit()