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

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.

This is very short answer.

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

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')

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)

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

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}")

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')

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)

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")

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')

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.

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

with open(str('file_name.jpg', 'wb') as handler:
    handler.write(img_data)

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

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)

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)