[python] How can I take a screenshot/image of a website using Python?

What I want to achieve is to get a website screenshot from any website in python.

Env: Linux

You don't mention what environment you're running in, which makes a big difference because there isn't a pure Python web browser that's capable of rendering HTML.

But if you're using a Mac, I've used webkit2png with great success. If not, as others have pointed out there are plenty of options.

import subprocess

def screenshots(url, name):
    subprocess.run('webkit2png -F -o {} {} -D ./screens'.format(name, url), 
      shell=True)

Here is my solution by grabbing help from various sources. It takes full web page screen capture and it crops it (optional) and generates thumbnail from the cropped image also. Following are the requirements:

Requirements:

1. Install NodeJS
2. Using Node's package manager install phantomjs: npm -g install phantomjs
3. Install selenium (in your virtualenv, if you are using that)
4. Install imageMagick
5. Add phantomjs to system path (on windows)

import os
from subprocess import Popen, PIPE
from selenium import webdriver

abspath = lambda *p: os.path.abspath(os.path.join(*p))
ROOT = abspath(os.path.dirname(__file__))


def execute_command(command):
    result = Popen(command, shell=True, stdout=PIPE).stdout.read()
    if len(result) > 0 and not result.isspace():
        raise Exception(result)


def do_screen_capturing(url, screen_path, width, height):
    print "Capturing screen.."
    driver = webdriver.PhantomJS()
    # it save service log file in same directory
    # if you want to have log file stored else where
    # initialize the webdriver.PhantomJS() as
    # driver = webdriver.PhantomJS(service_log_path='/var/log/phantomjs/ghostdriver.log')
    driver.set_script_timeout(30)
    if width and height:
        driver.set_window_size(width, height)
    driver.get(url)
    driver.save_screenshot(screen_path)


def do_crop(params):
    print "Croping captured image.."
    command = [
        'convert',
        params['screen_path'],
        '-crop', '%sx%s+0+0' % (params['width'], params['height']),
        params['crop_path']
    ]
    execute_command(' '.join(command))


def do_thumbnail(params):
    print "Generating thumbnail from croped captured image.."
    command = [
        'convert',
        params['crop_path'],
        '-filter', 'Lanczos',
        '-thumbnail', '%sx%s' % (params['width'], params['height']),
        params['thumbnail_path']
    ]
    execute_command(' '.join(command))


def get_screen_shot(**kwargs):
    url = kwargs['url']
    width = int(kwargs.get('width', 1024)) # screen width to capture
    height = int(kwargs.get('height', 768)) # screen height to capture
    filename = kwargs.get('filename', 'screen.png') # file name e.g. screen.png
    path = kwargs.get('path', ROOT) # directory path to store screen

    crop = kwargs.get('crop', False) # crop the captured screen
    crop_width = int(kwargs.get('crop_width', width)) # the width of crop screen
    crop_height = int(kwargs.get('crop_height', height)) # the height of crop screen
    crop_replace = kwargs.get('crop_replace', False) # does crop image replace original screen capture?

    thumbnail = kwargs.get('thumbnail', False) # generate thumbnail from screen, requires crop=True
    thumbnail_width = int(kwargs.get('thumbnail_width', width)) # the width of thumbnail
    thumbnail_height = int(kwargs.get('thumbnail_height', height)) # the height of thumbnail
    thumbnail_replace = kwargs.get('thumbnail_replace', False) # does thumbnail image replace crop image?

    screen_path = abspath(path, filename)
    crop_path = thumbnail_path = screen_path

    if thumbnail and not crop:
        raise Exception, 'Thumnail generation requires crop image, set crop=True'

    do_screen_capturing(url, screen_path, width, height)

    if crop:
        if not crop_replace:
            crop_path = abspath(path, 'crop_'+filename)
        params = {
            'width': crop_width, 'height': crop_height,
            'crop_path': crop_path, 'screen_path': screen_path}
        do_crop(params)

        if thumbnail:
            if not thumbnail_replace:
                thumbnail_path = abspath(path, 'thumbnail_'+filename)
            params = {
                'width': thumbnail_width, 'height': thumbnail_height,
                'thumbnail_path': thumbnail_path, 'crop_path': crop_path}
            do_thumbnail(params)
    return screen_path, crop_path, thumbnail_path


if __name__ == '__main__':
    '''
        Requirements:
        Install NodeJS
        Using Node's package manager install phantomjs: npm -g install phantomjs
        install selenium (in your virtualenv, if you are using that)
        install imageMagick
        add phantomjs to system path (on windows)
    '''

    url = 'http://stackoverflow.com/questions/1197172/how-can-i-take-a-screenshot-image-of-a-website-using-python'
    screen_path, crop_path, thumbnail_path = get_screen_shot(
        url=url, filename='sof.png',
        crop=True, crop_replace=False,
        thumbnail=True, thumbnail_replace=False,
        thumbnail_width=200, thumbnail_height=150,
    )

These are the generated images:

• Full web page screen
• Cropped image from captured screen
• Thumbnail of a cropped image

can do using Selenium

from selenium import webdriver

DRIVER = 'chromedriver'
driver = webdriver.Chrome(DRIVER)
driver.get('https://www.spotify.com')
screenshot = driver.save_screenshot('my_screenshot.png')
driver.quit()

https://sites.google.com/a/chromium.org/chromedriver/getting-started

I created a library called pywebcapture that wraps selenium that will do just that:

pip install pywebcapture

Once you install with pip, you can do the following to easily get full size screenshots:

# import modules
from pywebcapture import loader, driver

# load csv with urls
csv_file = loader.CSVLoader("csv_file_with_urls.csv", has_header_bool, url_column, optional_filename_column)
uri_dict = csv_file.get_uri_dict()

# create instance of the driver and run
d = driver.Driver("path/to/webdriver/", output_filepath, delay, uri_dict)
d.run()

Enjoy!

https://pypi.org/project/pywebcapture/

Here is a simple solution using webkit: http://webscraping.com/blog/Webpage-screenshots-with-webkit/

import sys
import time
from PyQt4.QtCore import *
from PyQt4.QtGui import *
from PyQt4.QtWebKit import *

class Screenshot(QWebView):
    def __init__(self):
        self.app = QApplication(sys.argv)
        QWebView.__init__(self)
        self._loaded = False
        self.loadFinished.connect(self._loadFinished)

    def capture(self, url, output_file):
        self.load(QUrl(url))
        self.wait_load()
        # set to webpage size
        frame = self.page().mainFrame()
        self.page().setViewportSize(frame.contentsSize())
        # render image
        image = QImage(self.page().viewportSize(), QImage.Format_ARGB32)
        painter = QPainter(image)
        frame.render(painter)
        painter.end()
        print 'saving', output_file
        image.save(output_file)

    def wait_load(self, delay=0):
        # process app events until page loaded
        while not self._loaded:
            self.app.processEvents()
            time.sleep(delay)
        self._loaded = False

    def _loadFinished(self, result):
        self._loaded = True

s = Screenshot()
s.capture('http://webscraping.com', 'website.png')
s.capture('http://webscraping.com/blog', 'blog.png')

Using a web service s-shot.ru (so it's not so fast), but quite easy to set up what need through the link configuration. And you can easily capture full page screenshots

import requests
import urllib.parse

BASE = 'https://mini.s-shot.ru/1024x0/JPEG/1024/Z100/?' # you can modify size, format, zoom
url = 'https://stackoverflow.com/'#or whatever link you need
url = urllib.parse.quote_plus(url) #service needs link to be joined in encoded format
print(url)

path = 'target1.jpg'
response = requests.get(BASE + url, stream=True)

if response.status_code == 200:
    with open(path, 'wb') as file:
        for chunk in response:
            file.write(chunk)

You can use Google Page Speed API to achieve your task easily. In my current project, I have used Google Page Speed API`s query written in Python to capture screenshots of any Web URL provided and save it to a location. Have a look.

import urllib2
import json
import base64
import sys
import requests
import os
import errno

#   The website's URL as an Input
site = sys.argv[1]
imagePath = sys.argv[2]

#   The Google API.  Remove "&strategy=mobile" for a desktop screenshot
api = "https://www.googleapis.com/pagespeedonline/v1/runPagespeed?screenshot=true&strategy=mobile&url=" + urllib2.quote(site)

#   Get the results from Google
try:
    site_data = json.load(urllib2.urlopen(api))
except urllib2.URLError:
    print "Unable to retreive data"
    sys.exit()

try:
    screenshot_encoded =  site_data['screenshot']['data']
except ValueError:
    print "Invalid JSON encountered."
    sys.exit()

#   Google has a weird way of encoding the Base64 data
screenshot_encoded = screenshot_encoded.replace("_", "/")
screenshot_encoded = screenshot_encoded.replace("-", "+")

#   Decode the Base64 data
screenshot_decoded = base64.b64decode(screenshot_encoded)

if not os.path.exists(os.path.dirname(impagepath)):
    try:
        os.makedirs(os.path.dirname(impagepath))
        except  OSError as exc:
            if exc.errno  != errno.EEXIST:
                raise

#   Save the file
with open(imagePath, 'w') as file_:
    file_.write(screenshot_decoded)

Unfortunately, following are the drawbacks. If these do not matter, you can proceed with Google Page Speed API. It works well.

• The maximum width is 320px
• According to Google API Quota, there is a limit of 25,000 requests per day

This is an old question and most answers are a bit dated. Currently, I would do 1 of 2 things.

1. Create a program that takes the screenshots

I would use Pyppeteer to take screenshots of websites. This runs on the Puppeteer package. Puppeteer spins up a headless chrome browser, so the screenshots will look exactly like they would in a normal browser.

This is taken from the pyppeteer documentation:

import asyncio
from pyppeteer import launch

async def main():
    browser = await launch()
    page = await browser.newPage()
    await page.goto('https://example.com')
    await page.screenshot({'path': 'example.png'})
    await browser.close()

asyncio.get_event_loop().run_until_complete(main())

2. Use a screenshot API

You could also use a screenshot API such as this one. The nice thing is that you don't have to set everything up yourself but can simply call an API endpoint.

This is taken from the screenshot API's documentation:

import urllib.parse
import urllib.request
import ssl

ssl._create_default_https_context = ssl._create_unverified_context

# The parameters.
token = "YOUR_API_TOKEN"
url = urllib.parse.quote_plus("https://example.com")
width = 1920
height = 1080
output = "image"

# Create the query URL.
query = "https://screenshotapi.net/api/v1/screenshot"
query += "?token=%s&url=%s&width=%d&height=%d&output=%s" % (token, url, width, height, output)

# Call the API.
urllib.request.urlretrieve(query, "./example.png")

Try this..

#!/usr/bin/env python

import gtk.gdk

import time

import random

while 1 :
    # generate a random time between 120 and 300 sec
    random_time = random.randrange(120,300)

    # wait between 120 and 300 seconds (or between 2 and 5 minutes)
    print "Next picture in: %.2f minutes" % (float(random_time) / 60)

    time.sleep(random_time)

    w = gtk.gdk.get_default_root_window()
    sz = w.get_size()

    print "The size of the window is %d x %d" % sz

    pb = gtk.gdk.Pixbuf(gtk.gdk.COLORSPACE_RGB,False,8,sz[0],sz[1])
    pb = pb.get_from_drawable(w,w.get_colormap(),0,0,0,0,sz[0],sz[1])

    ts = time.time()
    filename = "screenshot"
    filename += str(ts)
    filename += ".png"

    if (pb != None):
        pb.save(filename,"png")
        print "Screenshot saved to "+filename
    else:
        print "Unable to get the screenshot."

11 years later...
Taking a website screenshot using Python3.6 and Google PageSpeedApi Insights v5:

import base64
import requests
import traceback
import urllib.parse as ul

# It's possible to make requests without the api key, but the number of requests is very limited  

url = "https://duckgo.com"
urle = ul.quote_plus(url)
image_path = "duckgo.jpg"

key = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
strategy = "desktop" # "mobile"
u = f"https://www.googleapis.com/pagespeedonline/v5/runPagespeed?key={key}&strategy={strategy}&url={urle}"

try:
    j = requests.get(u).json()
    ss_encoded = j['lighthouseResult']['audits']['final-screenshot']['details']['data'].replace("data:image/jpeg;base64,", "")
    ss_decoded = base64.b64decode(ss_encoded)
    with open(image_path, 'wb+') as f:
        f.write(ss_decoded) 
except :
    print(traceback.format_exc())
    exit(1)

Notes:

• Live Demo
• Pros: Free
• Conns: Low Resolution
• Get API Key
• Docs
• Limits:
  • Queries per day = 25,000
  • Queries per 100 seconds = 400

I can't comment on ars's answer, but I actually got Roland Tapken's code running using QtWebkit and it works quite well.

Just wanted to confirm that what Roland posts on his blog works great on Ubuntu. Our production version ended up not using any of what he wrote but we are using the PyQt/QtWebKit bindings with much success.

Note: The URL used to be: http://www.blogs.uni-osnabrueck.de/rotapken/2008/12/03/create-screenshots-of-a-web-page-using-python-and-qtwebkit/ I've updated it with a working copy.

Using Rendertron is an option. Under the hood, this is a headless Chrome exposing the following endpoints:

• /render/:url: Access this route e.g. with requests.get if you are interested in the DOM.
• /screenshot/:url: Access this route if you are interested in a screenshot.

You would install rendertron with npm, run rendertron in one terminal, access http://localhost:3000/screenshot/:url and save the file, but a demo is available at render-tron.appspot.com making it possible to run this Python3 snippet locally without installing the npm package:

import requests

BASE = 'https://render-tron.appspot.com/screenshot/'
url = 'https://google.com'
path = 'target.jpg'
response = requests.get(BASE + url, stream=True)
# save file, see https://stackoverflow.com/a/13137873/7665691
if response.status_code == 200:
    with open(path, 'wb') as file:
        for chunk in response:
            file.write(chunk)