CODEWITHSUNDEEP
pythonurllib2python-requests
shkschneider
shkschneider

Reputation: 18243

How to download image using requests

I'm trying to download and save an image from the web using python's requests module.

Here is the (working) code I used:

img = urllib2.urlopen(settings.STATICMAP_URL.format(**data))
with open(path, 'w') as f:
    f.write(img.read())

Here is the new (non-working) code using requests:

r = requests.get(settings.STATICMAP_URL.format(**data))
if r.status_code == 200:
    img = r.raw.read()
    with open(path, 'w') as f:
        f.write(img)

Can you help me on what attribute from the response to use from requests?

Upvotes: 516

Views: 688293

Answers (19)

forzagreen
forzagreen

Reputation: 2683

To download the image from image_url to photo.jpg:

import requests
from pathlib import Path

Path("photo.jpg").write_bytes(requests.get(image_url).content)

Upvotes: 0

alvas
alvas

Reputation: 121992

TL;DR

Summarizing the great answers from others.

Method Needs requests Needs PIL Needs ...
requests.get -> shutil Yes No -
requests.get -> open(mode="wb") Yes No -
requests.get -> ByteIO -> Image.save Yes Yes -
urllib - - -
wget No No wget
requests.get -> PIL.Image -> np.save Yes Yes numpy

Use shutil and output the decoded raw content from requests.get

Original answer modified from https://stackoverflow.com/a/13137873/610569

import shutil
import requests

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

response = requests.get(img_url, stream=True)        
with open('dpreview.jpg', 'wb') as fout:
    response.raw.decode_content = True
    shutil.copyfileobj(response.raw, fout)

Writing the binary directly into file I/O

import requests

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

response = requests.get(img_url, stream=True) 

with open('dpreview.jpg', 'wb') as fout:
    for chunk in response:
        fout.write(chunk)

Stream content into io.BytesIO into PIL.Image object and save it

from io import BytesIO

import requests
from PIL import Image

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

# Stream to BytesIO
response = requests.get(img_url, stream=True)
img = Image.open(BytesIO(response.content))
img.save('dpreview.jpg')


# Using raw content
response = requests.get(img_url, stream=True)
img = Image.open(response.raw)
img.save('dpreview.jpg')

Use urllib

Original answer from https://stackoverflow.com/a/33866125/610569

import urllib

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

urllib.request.urlretrieve(img_url, "dpreview.jpg")

And if a specific user-agent is needed for the request, from https://stackoverflow.com/a/69764951/610569

import urllib

opener=urllib.request.build_opener()
opener.addheaders=[('User-Agent','Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.102 Safari/537.36 Edge/18.19582')]
urllib.request.install_opener(opener)

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

urllib.request.urlretrieve(img_url, "dpreview.jpg")

Use wget

import wget

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

wget.download(img_url, out='dpreview.jpg')

Saving PIL.Image as numpy array

import requests
from PIL import Image

import numpy as np


img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

response = requests.get(img_url, stream=True) 
img = Image.open(response.raw)

# Converts and save image into numpy array.
np.save('dpreview.npy', np.asarray(img))

# Loads a npy file to Image
img_arr = np.load('dpreview.npy')
img = Image.fromarray(img_arr.astype(np.uint8))

Upvotes: 8

Zhenyi Zhang
Zhenyi Zhang

Reputation: 1295

I have the same need for downloading images using requests. I first tried the answer of Martijn Pieters, and it works well. But when I did a profile on this simple function, I found that it uses so many function calls compared to urllib and urllib2.

I then tried the way recommended by the author of requests module:

import requests
from PIL import Image
# python2.x, use this instead  
# from StringIO import StringIO
# for python3.x,
from io import StringIO

r = requests.get('https://example.com/image.jpg')
i = Image.open(StringIO(r.content))

This much more reduced the number of function calls, thus speeded up my application. Here is the code of my profiler and the result.

#!/usr/bin/python
import requests
from StringIO import StringIO
from PIL import Image
import profile

def testRequest():
    image_name = 'test1.jpg'
    url = 'http://example.com/image.jpg'

    r = requests.get(url, stream=True)
    with open(image_name, 'wb') as f:
        for chunk in r.iter_content():
            f.write(chunk)

def testRequest2():
    image_name = 'test2.jpg'
    url = 'http://example.com/image.jpg'

    r = requests.get(url)
    
    i = Image.open(StringIO(r.content))
    i.save(image_name)

if __name__ == '__main__':
    profile.run('testUrllib()')
    profile.run('testUrllib2()')
    profile.run('testRequest()')

The result for testRequest:

343080 function calls (343068 primitive calls) in 2.580 seconds

And the result for testRequest2:

3129 function calls (3105 primitive calls) in 0.024 seconds

Upvotes: 91

unofficialdxnny
unofficialdxnny

Reputation: 311

Here is a very simple code

import requests

response = requests.get("https://i.imgur.com/ExdKOOz.png") ## Making a variable to get image.

file = open("sample_image.png", "wb") ## Creates the file for image
file.write(response.content) ## Saves file content
file.close()

Upvotes: -1

Adriano_Pinaffo
Adriano_Pinaffo

Reputation: 1699

my approach was to use response.content (blob) and save to the file in binary mode

img_blob = requests.get(url, timeout=5).content
with open(destination + '/' + title, 'wb') as img_file:
     img_file.write(img_blob)

Check out my python project that downloads images from unsplash.com based on keywords.

Upvotes: 5

Dmitriy Zub
Dmitriy Zub

Reputation: 1724

Agree with Blairg23 that using urllib.request.urlretrieve is one of the easiest solutions.

One note I want to point out here. Sometimes it won't download anything because the request was sent via script (bot), and if you want to parse images from Google images or other search engines, you need to pass user-agent to request headers first, and then download the image, otherwise, the request will be blocked and it will throw an error.

Pass user-agent and download image:

opener=urllib.request.build_opener()
opener.addheaders=[('User-Agent','Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.102 Safari/537.36 Edge/18.19582')]
urllib.request.install_opener(opener)

urllib.request.urlretrieve(URL, 'image_name.jpg')

Code in the online IDE that scrapes and downloads images from Google images using requests, bs4, urllib.requests.

Alternatively, if your goal is to scrape images from search engines like Google, Bing, Yahoo!, DuckDuckGo (and other search engines), then you can use SerpApi. It's a paid API with a free plan.

The biggest difference is that there's no need to figure out how to bypass blocks from search engines or how to extract certain parts from the HTML or JavaScript since it's already done for the end-user.

Example code to integrate:

import os, urllib.request
from serpapi import GoogleSearch

params = {
  "api_key": os.getenv("API_KEY"),
  "engine": "google",
  "q": "pexels cat",
  "tbm": "isch"
}

search = GoogleSearch(params)
results = search.get_dict()

print(json.dumps(results['images_results'], indent=2, ensure_ascii=False))

# download images 
for index, image in enumerate(results['images_results']):

    # print(f'Downloading {index} image...')
    
    opener=urllib.request.build_opener()
    opener.addheaders=[('User-Agent','Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.102 Safari/537.36 Edge/18.19582')]
    urllib.request.install_opener(opener)

    # saves original res image to the SerpApi_Images folder and add index to the end of file name
    urllib.request.urlretrieve(image['original'], f'SerpApi_Images/original_size_img_{index}.jpg')

-----------
'''
]
  # other images
  {
    "position": 100, # 100 image
    "thumbnail": "https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQK62dIkDjNCvEgmGU6GGFZcpVWwX-p3FsYSg&usqp=CAU",
    "source": "homewardboundnj.org",
    "title": "pexels-helena-lopes-1931367 - Homeward Bound Pet Adoption Center",
    "link": "https://homewardboundnj.org/upcoming-event/black-cat-appreciation-day/pexels-helena-lopes-1931367/",
    "original": "https://homewardboundnj.org/wp-content/uploads/2020/07/pexels-helena-lopes-1931367.jpg",
    "is_product": false
  }
]
'''

Disclaimer, I work for SerpApi.

Upvotes: 0

David Johnson
David Johnson

Reputation: 9

for download Image

import requests
Picture_request = requests.get(url)

Upvotes: -3

Blairg23
Blairg23

Reputation: 12045

This might be easier than using requests. This is the only time I'll ever suggest not using requests to do HTTP stuff.

Two liner using urllib:

>>> import urllib
>>> urllib.request.urlretrieve("http://www.example.com/songs/mp3.mp3", "mp3.mp3")

There is also a nice Python module named wget that is pretty easy to use. Found here.

This demonstrates the simplicity of the design:

>>> import wget
>>> url = 'http://www.futurecrew.com/skaven/song_files/mp3/razorback.mp3'
>>> filename = wget.download(url)
100% [................................................] 3841532 / 3841532>
>> filename
'razorback.mp3'

Enjoy.

Edit: You can also add an out parameter to specify a path.

>>> out_filepath = <output_filepath>    
>>> filename = wget.download(url, out=out_filepath)

Upvotes: 74

Martijn Pieters
Martijn Pieters

Reputation: 1121226

You can either use the response.raw file object, or iterate over the response.

To use the response.raw file-like object will not, by default, decode compressed responses (with GZIP or deflate). You can force it to decompress for you anyway by setting the decode_content attribute to True (requests sets it to False to control decoding itself). You can then use shutil.copyfileobj() to have Python stream the data to a file object:

import requests
import shutil

r = requests.get(settings.STATICMAP_URL.format(**data), stream=True)
if r.status_code == 200:
    with open(path, 'wb') as f:
        r.raw.decode_content = True
        shutil.copyfileobj(r.raw, f)

To iterate over the response use a loop; iterating like this ensures that data is decompressed by this stage:

r = requests.get(settings.STATICMAP_URL.format(**data), stream=True)
if r.status_code == 200:
    with open(path, 'wb') as f:
        for chunk in r:
            f.write(chunk)

This'll read the data in 128 byte chunks; if you feel another chunk size works better, use the Response.iter_content() method with a custom chunk size:

r = requests.get(settings.STATICMAP_URL.format(**data), stream=True)
if r.status_code == 200:
    with open(path, 'wb') as f:
        for chunk in r.iter_content(1024):
            f.write(chunk)

Note that you need to open the destination file in binary mode to ensure python doesn't try and translate newlines for you. We also set stream=True so that requests doesn't download the whole image into memory first.

Upvotes: 658

Katja S&#252;ss
Katja S&#252;ss

Reputation: 759

Following code snippet downloads a file.

The file is saved with its filename as in specified url.

import requests

url = "http://example.com/image.jpg"
filename = url.split("/")[-1]
r = requests.get(url, timeout=0.5)

if r.status_code == 200:
    with open(filename, 'wb') as f:
        f.write(r.content)

Upvotes: 42

Harshit Singhai
Harshit Singhai

Reputation: 1280

This is how I did it

import requests
from PIL import Image
from io import BytesIO

url = 'your_url'
files = {'file': ("C:/Users/shadow/Downloads/black.jpeg", open('C:/Users/shadow/Downloads/black.jpeg', 'rb'),'image/jpg')}
response = requests.post(url, files=files)

img = Image.open(BytesIO(response.content))
img.show()

Upvotes: 6

duhaime
duhaime

Reputation: 27574

This is the first response that comes up for google searches on how to download a binary file with requests. In case you need to download an arbitrary file with requests, you can use:

import requests
url = 'https://s3.amazonaws.com/lab-data-collections/GoogleNews-vectors-negative300.bin.gz'
open('GoogleNews-vectors-negative300.bin.gz', 'wb').write(requests.get(url, allow_redirects=True).content)

Upvotes: 5

Jyotiprakash Das
Jyotiprakash Das

Reputation: 397

You can do something like this:

import requests
import random

url = "https://images.pexels.com/photos/1308881/pexels-photo-1308881.jpeg? auto=compress&cs=tinysrgb&dpr=1&w=500"
name=random.randrange(1,1000)
filename=str(name)+".jpg"
response = requests.get(url)
if response.status_code.ok:
   with open(filename,'w') as f:
    f.write(response.content)

Upvotes: 1

Riccardo D
Riccardo D

Reputation: 621

As easy as to import Image and requests

from PIL import Image
import requests

img = Image.open(requests.get(url, stream = True).raw)
img.save('img1.jpg')

Upvotes: 19

kiranbkrishna
kiranbkrishna

Reputation: 2580

How about this, a quick solution.

import requests

url = "http://craphound.com/images/1006884_2adf8fc7.jpg"
response = requests.get(url)
if response.status_code == 200:
    with open("/Users/apple/Desktop/sample.jpg", 'wb') as f:
        f.write(response.content)

Upvotes: 241

Wernight
Wernight

Reputation: 37600

There are 2 main ways:

  1. Using .content (simplest/official) (see Zhenyi Zhang's answer):

    import io  # Note: io.BytesIO is StringIO.StringIO on Python2.
import requests

r = requests.get('http://lorempixel.com/400/200')
r.raise_for_status()
with io.BytesIO(r.content) as f:
    with Image.open(f) as img:
        img.show()

  2. Using .raw (see Martijn Pieters's answer):

    import requests

r = requests.get('http://lorempixel.com/400/200', stream=True)
r.raise_for_status()
r.raw.decode_content = True  # Required to decompress gzip/deflate compressed responses.
with PIL.Image.open(r.raw) as img:
    img.show()
r.close()  # Safety when stream=True ensure the connection is released.

Timing both shows no noticeable difference.

Upvotes: 25

justincc
justincc

Reputation: 103

I'm going to post an answer as I don't have enough rep to make a comment, but with wget as posted by Blairg23, you can also provide an out parameter for the path.

 wget.download(url, out=path)

Upvotes: 5

Chris Redford
Chris Redford

Reputation: 17768

Here is a more user-friendly answer that still uses streaming.

Just define these functions and call getImage(). It will use the same file name as the url and write to the current directory by default, but both can be changed.

import requests
from StringIO import StringIO
from PIL import Image

def createFilename(url, name, folder):
    dotSplit = url.split('.')
    if name == None:
        # use the same as the url
        slashSplit = dotSplit[-2].split('/')
        name = slashSplit[-1]
    ext = dotSplit[-1]
    file = '{}{}.{}'.format(folder, name, ext)
    return file

def getImage(url, name=None, folder='./'):
    file = createFilename(url, name, folder)
    with open(file, 'wb') as f:
        r = requests.get(url, stream=True)
        for block in r.iter_content(1024):
            if not block:
                break
            f.write(block)

def getImageFast(url, name=None, folder='./'):
    file = createFilename(url, name, folder)
    r = requests.get(url)
    i = Image.open(StringIO(r.content))
    i.save(file)

if __name__ == '__main__':
    # Uses Less Memory
    getImage('http://www.example.com/image.jpg')
    # Faster
    getImageFast('http://www.example.com/image.jpg')

The request guts of getImage() are based on the answer here and the guts of getImageFast() are based on the answer above.

Upvotes: 5

Oleh Prypin
Oleh Prypin

Reputation: 34116

Get a file-like object from the request and copy it to a file. This will also avoid reading the whole thing into memory at once.

import shutil

import requests

url = 'http://example.com/img.png'
response = requests.get(url, stream=True)
with open('img.png', 'wb') as out_file:
    shutil.copyfileobj(response.raw, out_file)
del response

Upvotes: 303

Related Questions