Develop Reference

Getting a thumbnail image from google books api into a python variable

I can do this OK both in js and php but not in python. I'm trying to pull a thumbnail image from google books api into a python variable.
The text objects are fine eg
newTitle = (parsed_json['items'][0]['volumeInfo']['title'])
isbn10 = (parsed_json['items'][0]['volumeInfo']['industryIdentifiers'][1]['identifier'])
isbn13 = (parsed_json['items'][0]['volumeInfo']['industryIdentifiers'][0]['identifier'])
The image is supplied in the api as follows. (if you put the http// url into a browser you see the image):
"imageLinks": {
"smallThumbnail": "http://books.google.com/books/content?id=XUnNDwAAQBAJ&printsec=frontcover&img=1&zoom=5&edge=curl&source=gbs_api",
"thumbnail": "http://books.google.com/books/content?id=XUnNDwAAQBAJ&printsec=frontcover&img=1&zoom=1&edge=curl&source=gbs_api"
I have tried the simple:
myImage = (parsed_json['items'][0]['volumeInfo']['imageLinks'][thumbnail])
which doesn't work.
I have installed pillow to provide image management:
from PIL import Image
img = Image.open("parsed_json['items'][0]['volumeInfo']['imageLinks'][thumbnail]") or
img = Image.open(parsed_json['items'][0]['volumeInfo']['imageLinks'][thumbnail])
which doesn't work. I have tried more complex arrangements:
from PIL import Image
import requests
from io import BytesIO
response = requests.get(parsed_json['items'][0]['volumeInfo']['imageLinks'][thumbnail])
img = Image.open(BytesIO(response.content))
but nothing seems to work. I have tried many other variations on these attempts. I have also, unsuccessfully tried to load the text that points to the thumbnail to try another route. I am confident that the "['items'][0]['volumeInfo']['imageLinks'][thumbnail]" is correct though my only way of testing whether the variable is properly loaded is to save it or if the line of code isn't working.

I didn't have problems downloading and opening the image.
I have use the following code
import requests
from PIL import Image
image_url = "https://books.google.com/books/content?id=XUnNDwAAQBAJ&printsec=frontcover&img=1&zoom=5&edge=curl&source=gbs_api"
r = requests.get(image_url)
with open("demo_image",'wb') as f:
f.write(r.content)
img = Image.open("demo_image")

recommended way to download images in python requests

I see that there are two ways to download images using python-reuqests.
Uisng PIL as stated in docs (https://requests.readthedocs.io/en/master/user/quickstart/#binary-response-content):
from PIL import Image
from io import BytesIO
i = Image.open(BytesIO(r.content))
using streamed response content:
r = requests.get(url, stream=True)
with open(image_name, 'wb') as f:
for chunk in r.iter_content():
f.write(chunk)
Which is the recommended wya to download images however? both have its merits I suyppose, and I was wondering what is the optimal approach.

I love the minimalist way. There is nothing called right way. It depends on the task you want to perform and the constraints you have.
import requests
with open('file.png', 'wb') as f:
f.write(requests.get(url).content)
# if you change png to jpg, there will be no error

I did use the below lines of code in a function to save images.
# import the required libraries from Python
import pathlib,urllib.request,os,uuid
# URL of the image you want to download
image_url = "https://example.com/image.png"
# Using the uuid generate new and unique names for your images
filename = str(uuid.uuid4())
# Strip the image extension from it's original name
file_ext = pathlib.Path(image_url).suffix
# Join the new image name to the extension
picture_filename = filename + file_ext
# 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, picture_filename)
# Using "urlretrieve()" from urllib.request save the image
urllib.request.urlretrieve(image_url, 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

Image processing after upload with Python Bottle

Context
I have made a simple web app for uploading content to a blog. The front sends AJAX requests (using FormData) to the backend which is Bottle running on Python 3.7. Text content is saved to a MySQL database and images are saved to a folder on the server. Everything works fine.
Image processing and PIL/Pillow
Now, I want to enable processing of uploaded images to standardise them (I need them all resized and/or cropped to 700x400px).
I was hoping to use Pillow for this. My problem is creating a PIL Image object from the file object in Bottle. I cannot initialise a valid Image object.
Code
# AJAX sends request to this route
#post('/update')
def update():
# Form data
title = request.forms.get("title")
body = request.forms.get("body")
image = request.forms.get("image")
author = request.forms.get("author")
# Image upload
file = request.files.get("file")
if file:
extension = file.filename.split(".")[-1]
if extension not in ('png', 'jpg', 'jpeg'):
return {"result" : 0, "message": "File Format Error"}
save_path = "my/save/path"
file.save(save_path)
The problem
This all works as expected, but I cannot create a valid Image object with pillow for processing. I even tried reloading the saved image using the save path but this did not work either.
Other attempts
The code below did not work. It caused an internal server error, though I am having trouble setting up more detailed Python debugging.
path = save_path + "/" + file.filename
image_data = open(path, "rb")
image = Image.open(image_data)
When logged manually, the path is a valid relative URL ("../domain-folder/images") and I have checked that I am definitely importing PIL (Pillow) correctly using PIL.PILLOW_VERSION.
I tried adapting this answer:
image = Image.frombytes('RGBA', (128,128), image_data, 'raw')
However, I won’t know the size until I have created the Image object. I also tried using io:
image = Image.open(io.BytesIO(image_data))
This did not work either. In each case, it is only the line trying to initialise the Image object that causes problems.
Summary
The Bottle documentation says the uploaded file is a file-like object, but I am not having much success in creating an Image object that I can process.
How should I go about this? I do not have a preference about processing before or after saving. I am comfortable with the processing, it is initialising the Image object that is causing the problem.
Edit - Solution
I got this to work by adapting the answer from eatmeimadanish. I had to use a io.BytesIO object to save the file from Bottle, then load it with Pillow from there. After processing, it could be saved in the usual way.
obj = io.BytesIO()
file.save(obj) # This saves the file retrieved by Bottle to the BytesIO object
path = save_path + "/" + file.filename
# Image processing
im = Image.open(obj) # Reopen the object with PIL
im = im.resize((700,400))
im.save(path, optimize=True)
I found this from the Pillow documentation about a different function that may also be of use.
PIL.Image.frombuffer(mode, size, data, decoder_name='raw', *args)
Note that this function decodes pixel data only, not entire images.
If you have an entire image file in a string, wrap it in a BytesIO object, and use open() to load it.

Use StringIO instead.
From PIL import Image
try:
import cStringIO as StringIO
except ImportError:
import StringIO
s = StringIO.StringIO()
#save your in memory file to this instead of a regular file
file = request.files.get("file")
if file:
extension = file.filename.split(".")[-1]
if extension not in ('png', 'jpg', 'jpeg'):
return {"result" : 0, "message": "File Format Error"}
file.save(s)
im = Image.open(s)
im.resize((700,400))
im.save(s, 'png', optimize=True)
s64 = base64.b64encode(s.getvalue())

From what I understand, you're trying to resize the image after it has been saved locally (note that you could try to do the resize before it is saved). If this is what you want to achieve here, you can open the image directly using Pillow, it does the job for you (you do not have to open(path, "rb"):
image = Image.open(path)
image.resize((700,400)).save(path)

How to get image from response and save it in models.ImageField

Asking this I realize that topic is quite common and perhaps was discussed many times. But it still not clear for me, so I'm trying to get image file from response, resize it and save it in model. I deal with the error:
AttributeError at /saveimage/
'InMemoryUploadedFile' object has no attribute 'get'
my code is like so:
import Image
import cStringIO
from django.core.files.uploadedfile import InMemoryUploadedFile
def get_and_save_image():
file = cStringIO.StringIO()
size = (200,200)
color = (255,0,0,0)
image = Image.new("RGBA", size, color)
image.save(file, format='JPEG')
image_file = InMemoryUploadedFile(file, None, 'foo.jpg', 'jpeg', None, None)
image_file.seek(0)
return image_file
Http response:
FILES:
some_image = <InMemoryUploadedFile: image.jpg (image/jpeg)>
Is it the proper way to save image into ImageField? Thanks in advance.

Here are two ways you can do this:
Save the image in a models.ImageField, and then return the image's url in the response, and you can create the image element in javascript and add the url.
Encode the image's data in base64 and return it as a string, which you can easily use in the front end. Here's an example of this:
Backend:
from PIL import Image
from io import BytesIO
import base64
def get_and_save_image():
size = (200,200)
color = (255,0,0,0)
image = Image.new("RGBA", size, color)
img_buffer = BytesIO()
image.save(img_buffer, format='JPEG')
response_string = base64.b64encode(img_buffer.getvalue())
return response_string
Front end:
<img src="data:image/jpg;base64,{{ THE_RESPONSE_STRING }}">

Serve a dynamically generated image with Django

How do I serve a dynamically generated image in Django?
I have an html tag
<html>
...
<img src="images/dynamic_chart.png" />
...
</html>
linked up to this request handler, which creates an in-memory image
def chart(request):
img = Image.new("RGB", (300,300), "#FFFFFF")
data = [(i,randint(100,200)) for i in range(0,300,10)]
draw = ImageDraw.Draw(img)
draw.polygon(data, fill="#000000")
# now what?
return HttpResponse(output)
I also plan to change the requests to AJAX, and add some sort of caching mechanism, but my understanding is that wouldn't affect this part of the solution.

I assume you're using PIL (Python Imaging Library). You need to replace your last line with (for example, if you want to serve a PNG image):
response = HttpResponse(mimetype="image/png")
img.save(response, "PNG")
return response
See here for more information.

I'm relatively new to Django myself. I haven't been able to find anything in Django itself, but I have stumbled upon a project on Google Code that may be of some help to you:
django-dynamic-media-serve

I was looking for a solution of the same problem
And for me this simple approach worked fine:
from django.http import FileResponse
def dyn_view(request):
response = FileResponse(open("image.png","rb"))
return response

Another way is to use BytesIO. BytesIO is like a buffer. So one can save the image (which is fast enough than writing to disk) in that buffer.
from PIL import Image, ImageDraw
import io
def chart(request):
img = Image.new('RGB', (240, 240), color=(250,160,170))
draw = ImageDraw.Draw(img)
draw.text((20, 40), 'some_text')
buff = io.BytesIO()
img.save(buff, 'jpeg')
return HttpResponse(buff.getvalue(), content_type='image/jpeg')