How to save an image locally using Python whose URL address I already know?

Python Problem Overview

I know the URL of an image on Internet.

e.g. http://www.digimouth.com/news/media/2011/09/google-logo.jpg, which contains the logo of Google.

Now, how can I download this image using Python without actually opening the URL in a browser and saving the file manually.

Python Solutions

Solution 1 - Python

Python 2

Here is a more straightforward way if all you want to do is save it as a file:

import urllib

urllib.urlretrieve("http://www.digimouth.com/news/media/2011/09/google-logo.jpg", "local-filename.jpg")

The second argument is the local path where the file should be saved.

Python 3

As SergO suggested the code below should work with Python 3.

import urllib.request

urllib.request.urlretrieve("http://www.digimouth.com/news/media/2011/09/google-logo.jpg", "local-filename.jpg")

Solution 2 - Python

import urllib
resource = urllib.urlopen("http://www.digimouth.com/news/media/2011/09/google-logo.jpg")
output = open("file01.jpg","wb")
output.write(resource.read())
output.close()

file01.jpg will contain your image.

Solution 3 - Python

I wrote a script that does just this, and it is available on my github for your use.

I utilized BeautifulSoup to allow me to parse any website for images. If you will be doing much web scraping (or intend to use my tool) I suggest you sudo pip install BeautifulSoup. Information on BeautifulSoup is available here.

For convenience here is my code:

from bs4 import BeautifulSoup
from urllib2 import urlopen
import urllib

# use this image scraper from the location that 
#you want to save scraped images to

def make_soup(url):
	html = urlopen(url).read()
	return BeautifulSoup(html)

def get_images(url):
	soup = make_soup(url)
	#this makes a list of bs4 element tags
	images = [img for img in soup.findAll('img')]
	print (str(len(images)) + "images found.")
	print 'Downloading images to current working directory.'
	#compile our unicode list of image links
	image_links = [each.get('src') for each in images]
	for each in image_links:
		filename=each.split('/')[-1]
		urllib.urlretrieve(each, filename)
	return image_links

#a standard call looks like this
#get_images('http://www.wookmark.com')

Solution 4 - Python

This can be done with requests. Load the page and dump the binary content to a file.

import os
import requests

url = 'https://apod.nasa.gov/apod/image/1701/potw1636aN159_HST_2048.jpg'
page = requests.get(url)

f_ext = os.path.splitext(url)[-1]
f_name = 'img{}'.format(f_ext)
with open(f_name, 'wb') as f:
    f.write(page.content)

Solution 5 - Python

Python 3

urllib.request — Extensible library for opening URLs

from urllib.error import HTTPError
from urllib.request import urlretrieve

try:
    urlretrieve(image_url, image_local_path)
except FileNotFoundError as err:
    print(err)   # something wrong with local path
except HTTPError as err:
    print(err)  # something wrong with url

Solution 6 - Python

A solution which works with Python 2 and Python 3:

try:
    from urllib.request import urlretrieve  # Python 3
except ImportError:
    from urllib import urlretrieve  # Python 2

url = "http://www.digimouth.com/news/media/2011/09/google-logo.jpg"
urlretrieve(url, "local-filename.jpg")

or, if the additional requirement of requests is acceptable and if it is a http(s) URL:

def load_requests(source_url, sink_path):
    """
    Load a file from an URL (e.g. http).

    Parameters
    ----------
    source_url : str
        Where to load the file from.
    sink_path : str
        Where the loaded file is stored.
    """
    import requests
    r = requests.get(source_url, stream=True)
    if r.status_code == 200:
        with open(sink_path, 'wb') as f:
            for chunk in r:
                f.write(chunk)

Solution 7 - Python

I made a script expanding on Yup.'s script. I fixed some things. It will now bypass 403:Forbidden problems. It wont crash when an image fails to be retrieved. It tries to avoid corrupted previews. It gets the right absolute urls. It gives out more information. It can be run with an argument from the command line.

# getem.py
# python2 script to download all images in a given url
# use: python getem.py http://url.where.images.are

from bs4 import BeautifulSoup
import urllib2
import shutil
import requests
from urlparse import urljoin
import sys
import time

def make_soup(url):
	req = urllib2.Request(url, headers={'User-Agent' : "Magic Browser"}) 
	html = urllib2.urlopen(req)
	return BeautifulSoup(html, 'html.parser')

def get_images(url):
	soup = make_soup(url)
	images = [img for img in soup.findAll('img')]
	print (str(len(images)) + " images found.")
	print 'Downloading images to current working directory.'
	image_links = [each.get('src') for each in images]
	for each in image_links:
		try:
			filename = each.strip().split('/')[-1].strip()
			src = urljoin(url, each)
			print 'Getting: ' + filename
			response = requests.get(src, stream=True)
			# delay to avoid corrupted previews
			time.sleep(1)
			with open(filename, 'wb') as out_file:
			    shutil.copyfileobj(response.raw, out_file)
		except:
			print '  An error occured. Continuing.'
	print 'Done.'

if __name__ == '__main__':
	url = sys.argv[1]
	get_images(url)

Solution 8 - Python

Using requests library

import requests
import shutil,os

headers = {
	'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/78.0.3904.108 Safari/537.36'
}
currentDir = os.getcwd()
path = os.path.join(currentDir,'Images')#saving images to Images folder

def ImageDl(url):
	attempts = 0
	while attempts < 5:#retry 5 times
		try:
			filename = url.split('/')[-1]
			r = requests.get(url,headers=headers,stream=True,timeout=5)
			if r.status_code == 200:
			    with open(os.path.join(path,filename),'wb') as f:
			        r.raw.decode_content = True
			        shutil.copyfileobj(r.raw,f)
			print(filename)
			break
		except Exception as e:
			attempts+=1
			print(e)


ImageDl(url)

Solution 9 - Python

This is very short answer.

import urllib
urllib.urlretrieve("http://photogallery.sandesh.com/Picture.aspx?AlubumId=422040", "Abc.jpg")

Solution 10 - Python

Use a simple python wget module to download the link. Usage below:

import wget
wget.download('http://www.digimouth.com/news/media/2011/09/google-logo.jpg')

Solution 11 - Python

Version for Python 3

I adjusted the code of @madprops for Python 3

# getem.py
# python2 script to download all images in a given url
# use: python getem.py http://url.where.images.are

from bs4 import BeautifulSoup
import urllib.request
import shutil
import requests
from urllib.parse import urljoin
import sys
import time

def make_soup(url):
    req = urllib.request.Request(url, headers={'User-Agent' : "Magic Browser"}) 
    html = urllib.request.urlopen(req)
    return BeautifulSoup(html, 'html.parser')

def get_images(url):
    soup = make_soup(url)
    images = [img for img in soup.findAll('img')]
    print (str(len(images)) + " images found.")
    print('Downloading images to current working directory.')
    image_links = [each.get('src') for each in images]
    for each in image_links:
        try:
            filename = each.strip().split('/')[-1].strip()
            src = urljoin(url, each)
            print('Getting: ' + filename)
            response = requests.get(src, stream=True)
            # delay to avoid corrupted previews
            time.sleep(1)
            with open(filename, 'wb') as out_file:
                shutil.copyfileobj(response.raw, out_file)
        except:
            print('  An error occured. Continuing.')
    print('Done.')

if __name__ == '__main__':
    get_images('http://www.wookmark.com')

Solution 12 - Python

Late answer, but for python>=3.6 you can use dload, i.e.:

import dload
dload.save("http://www.digimouth.com/news/media/2011/09/google-logo.jpg")

if you need the image as bytes, use:

img_bytes = dload.bytes("http://www.digimouth.com/news/media/2011/09/google-logo.jpg")

install using pip3 install dload

Solution 13 - Python

Something fresh for Python 3 using Requests:

Comments in the code. Ready to use function.

    
import requests
from os import path

def get_image(image_url):
    """
    Get image based on url.
    :return: Image name if everything OK, False otherwise
    """
    image_name = path.split(image_url)[1]
    try:
        image = requests.get(image_url)
    except OSError:  # Little too wide, but work OK, no additional imports needed. Catch all conection problems
        return False
    if image.status_code == 200:  # we could have retrieved error page
        base_dir = path.join(path.dirname(path.realpath(__file__)), "images") # Use your own path or "" to use current working directory. Folder must exist.
        with open(path.join(base_dir, image_name), "wb") as f:
            f.write(image.content)
        return image_name

get_image("https://apod.nasddfda.gov/apod/image/2003/S106_Mishra_1947.jpg")

Solution 14 - Python

this is the easiest method to download images.

import requests
from slugify import slugify

img_url = 'https://apod.nasa.gov/apod/image/1701/potw1636aN159_HST_2048.jpg'
img = requests.get(img_url).content
img_file = open(slugify(img_url) + '.' + str(img_url).split('.')[-1], 'wb')
img_file.write(img)
img_file.close()

Solution 15 - Python

If you don't already have the url for the image, you could scrape it with gazpacho:

from gazpacho import Soup
base_url = "http://books.toscrape.com"

soup = Soup.get(base_url)
links = [img.attrs["src"] for img in soup.find("img")]

And then download the asset with urllib as mentioned:

from pathlib import Path
from urllib.request import urlretrieve as download

directory = "images"
Path(directory).mkdir(exist_ok=True)

link = links[0]
name = link.split("/")[-1]

download(f"{base_url}/{link}", f"{directory}/{name}")

Solution 16 - Python

# import the required libraries from Python
import pathlib,urllib.request 

# Using pathlib, specify where the image is to be saved
downloads_path = str(pathlib.Path.home() / "Downloads")

# Form a full image path by joining the path to the 
# images' new name

picture_path  = os.path.join(downloads_path, "new-image.png")

# "/home/User/Downloads/new-image.png"

# Using "urlretrieve()" from urllib.request save the image 
urllib.request.urlretrieve("//example.com/image.png", picture_path)

# urlretrieve() takes in 2 arguments
# 1. The URL of the image to be downloaded
# 2. The image new name after download. By default, the image is saved
#    inside your current working directory

Solution 17 - Python

Download Image file, with avoiding all possible error:

import requests
import validators
from urllib.request import Request, urlopen
from urllib.error import URLError, HTTPError


def is_downloadable(url):
  valid=validators. url(url)
  if valid==False:
    return False
  req = Request(url)
  try:
    response = urlopen(req)
  except HTTPError as e:
    return False
  except URLError as e:
    return False
  else:
    return True



for i in range(len(File_data)):   #File data Contain list of address for image 
                                                      #file
  url = File_data[i][1]
  try:
    if (is_downloadable(url)):
      try:
        r = requests.get(url, allow_redirects=True)
        if url.find('/'):
          fname = url.rsplit('/', 1)[1]
          fname = pth+File_data[i][0]+"$"+fname #Destination to save 
                                                   #image file
          open(fname, 'wb').write(r.content)
      except Exception as e:
        print(e)
  except Exception as e:
    print(e)

Content Type	Original Author	Original Content on Stackoverflow
Question	Pankaj Vatsa	View Question on Stackoverflow
Solution 1 - Python	Liquid_Fire	View Answer on Stackoverflow
Solution 2 - Python	Noufal Ibrahim	View Answer on Stackoverflow
Solution 3 - Python	Yup.	View Answer on Stackoverflow
Solution 4 - Python	Alex	View Answer on Stackoverflow
Solution 5 - Python	SergO	View Answer on Stackoverflow
Solution 6 - Python	Martin Thoma	View Answer on Stackoverflow
Solution 7 - Python	madprops	View Answer on Stackoverflow
Solution 8 - Python	Sohan Das	View Answer on Stackoverflow
Solution 9 - Python	OO7	View Answer on Stackoverflow
Solution 10 - Python	Gaurav Shrivastava	View Answer on Stackoverflow
Solution 11 - Python	PythonProgrammi	View Answer on Stackoverflow
Solution 12 - Python	Pedro Lobito	View Answer on Stackoverflow
Solution 13 - Python	Pavel Pančocha	View Answer on Stackoverflow
Solution 14 - Python	Praveen Kumar	View Answer on Stackoverflow
Solution 15 - Python	emehex	View Answer on Stackoverflow
Solution 16 - Python	Dev Elie	View Answer on Stackoverflow
Solution 17 - Python	PRINCE ARYA	View Answer on Stackoverflow

How to save an image locally using Python whose URL address I already know?

Python Problem Overview

Python Solutions

Solution 1 - Python

Python 2

Python 3

Solution 2 - Python

Solution 3 - Python

Solution 4 - Python

Solution 5 - Python

Solution 6 - Python

Solution 7 - Python

Solution 8 - Python

Solution 9 - Python

Solution 10 - Python

Solution 11 - Python

Version for Python 3

Solution 12 - Python

Solution 13 - Python

Solution 14 - Python

Solution 15 - Python

Solution 16 - Python

Solution 17 - Python

Table Header Views in StoryBoards

FILE macro shows full path

Attributions

Python Problem Overview

Python Solutions

Solution 1 - Python

Python 2

Python 3

Solution 2 - Python

Solution 3 - Python

Solution 4 - Python

Solution 5 - Python

Solution 6 - Python

Solution 7 - Python

Solution 8 - Python

Solution 9 - Python

Solution 10 - Python

Solution 11 - Python

Version for Python 3

Solution 12 - Python

Solution 13 - Python

Solution 14 - Python

Solution 15 - Python

Solution 16 - Python

Solution 17 - Python

Table Header Views in StoryBoards

__FILE__ macro shows full path

Attributions

FILE macro shows full path