python save image from url

I got a problem when I am using python to save an image from url either by urllib2 request or urllib.urlretrieve. That is the url of the image is valid. I could download it manually using the explorer. However, when I use python to download the image, the file cannot be opened. I use Mac OS preview to view the image. Thank you!

UPDATE:

The code is as follow

def downloadImage(self):
    request = urllib2.Request(self.url)
    pic = urllib2.urlopen(request)
    print "downloading: " + self.url
    print self.fileName
    filePath = localSaveRoot + self.catalog  + self.fileName + Picture.postfix
    # urllib.urlretrieve(self.url, filePath)
    with open(filePath, 'wb') as localFile:
        localFile.write(pic.read())

The image URL that I want to download is http://site.meishij.net/r/58/25/3568808/a3568808_142682562777944.jpg

This URL is valid and I can save it through the browser but the python code would download a file that cannot be opened. The Preview says "It may be damaged or use a file format that Preview doesn't recognize." I compare the image that I download by Python and the one that I download manually through the browser. The size of the former one is several byte smaller. So it seems that the file is uncompleted, but I don't know why python cannot completely download it.

import requests

img_data = requests.get(image_url).content
with open('image_name.jpg', 'wb') as handler:
    handler.write(img_data)

A sample code that works for me on Windows:

import requests

with open('pic1.jpg', 'wb') as handle:
    response = requests.get(pic_url, stream=True)

    if not response.ok:
        print(response)

    for block in response.iter_content(1024):
        if not block:
            break

        handle.write(block)

It is the simplest way to download and save the image from internet using urlib.request package.

Here, you can simply pass the image URL(from where you want to download and save the image) and directory(where you want to save the download image locally, and give the image name with .jpg or .png) Here I given "local-filename.jpg" replace with this.

Python 3

import urllib.request
imgURL = "http://site.meishij.net/r/58/25/3568808/a3568808_142682562777944.jpg"

urllib.request.urlretrieve(imgURL, "D:/abc/image/local-filename.jpg")

You can download multiple images as well if you have all the image URLs from the internet. Just pass those image URLs in for loop, and the code automatically download the images from the internet.

Python code snippet to download a file from an url and save with its name

import requests

url = 'http://google.com/favicon.ico'
filename = url.split('/')[-1]
r = requests.get(url, allow_redirects=True)
open(filename, 'wb').write(r.content)

import random
import urllib.request

def download_image(url):
    name = random.randrange(1,100)
    fullname = str(name)+".jpg"
    urllib.request.urlretrieve(url,fullname)     
download_image("http://site.meishij.net/r/58/25/3568808/a3568808_142682562777944.jpg")

Anyone who is wondering how to get the image extension then you can try split method of string on image url:

str_arr = str(img_url).split('.')
img_ext = '.' + str_arr[3] #www.bigbasket.com/patanjali-atta.jpg (jpg is after 3rd dot so)
img_data = requests.get(img_url).content
with open(img_name + img_ext, 'wb') as handler:
    handler.write(img_data)

download and save image to directory

import requests

headers = {"User-Agent": "Mozilla/5.0 (X11; Linux x86_64; rv:60.0) Gecko/20100101 Firefox/60.0",
           "Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8",
           "Accept-Language": "en-US,en;q=0.9"
           }

img_data = requests.get(url=image_url, headers=headers).content
with open(create_dir() + "/" + 'image_name' + '.png', 'wb') as handler:
    handler.write(img_data)

for creating directory

def create_dir():
    # Directory
    dir_ = "CountryFlags"
    # Parent Directory path
    parent_dir = os.path.dirname(os.path.realpath(__file__))
    # Path
    path = os.path.join(parent_dir, dir_)
    os.mkdir(path)
    return path

For linux in case; you can use wget command

import os
url1 = 'YOUR_URL_WHATEVER'
os.system('wget {}'.format(url1))