urllib.request.urlopen Unknown file extension - python

I am trying to download a image from a URL using urllib and PIL. The url has no file extension, but the image is a jpg. I get a 'unknown file extension: ' error.
https://rimh2.domainstatic.com.au/afYrRIa5-BhhWW2t7eD7-cEOKro=/fit-in/1920x1080/filters:format(jpeg):quality(80):no_upscale()/http://b.domainstatic.com.au.s3-website-ap-southeast-2.amazonaws.com/2015800146_1_1_191119_062711-w800-h600
image = Image.open(urllib.request.urlopen(i))
image.save('path')

Try This Method
import urllib.request
urllib.request.urlretrieve("http://www.text.com/news/media/2019/12/text.jpg",
"image/filename.jpg")

If you want to save this file only, then go with #Bloody Programmer's answer but if you're going to process it as an image and just wanted to save it before processing, use
image.save('filename.jpg')
(Specify the extension)

from urllib import request
from PIL import Image
i = 'https://rimh2.domainstatic.com.au/afYrRIa5-BhhWW2t7eD7-cEOKro=/fit-in/1920x1080/filters:format(jpeg):quality(80):no_upscale()/http://b.domainstatic.com.au.s3-website-ap-southeast-2.amazonaws.com/2015800146_1_1_191119_062711-w800-h600'
img = Image.open(request.urlopen(i))
img = img.save('image.jpg')

Related

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")

loading png from json file using PIL

I am trying to use PIL.Image.open to open the image from an opened json file. Should I just put the name of the png file inside the Image.open() or is it the directory that leads to the json file?
Thanks
Here is a full example which is probably what you're looking for:
import json
from PIL import Image
with open('yourfilename.txt', 'r') as f:
data = json.load(f)
# Load the file path from the json
imgpath = data['yourkey']
# Place the image path into the open method
img = Image.open(imgpath)
The imgpath variable should return a string of the path to your image.
Pillows Image.open docs

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)

Entering the url in the browser, downloads the image automatically. How to write a python script to download such images?

url="https://images.data.gov.sg/api/traffic-images/2016/02/96128cfd-ab9a-4959-972e-a5e74bb149a9.jpg"
I am trying this:
import urllib
url="https://images.data.gov.sg/api/traffic-images/2016/02/96128cfd-ab9a-4959-972e-a5e74bb149a9.jpg"
IMAGE=url.rsplit("/")[-1]
urllib.urlretrieve(url,IMAGE)
Image is downloaded in the destination folder after the execution, but it is corrupt.
"Could not load image"; error pops up.
It might be because the domain that you are trying to reach has restrictions over download policy. Check this one out, hope it helps! https://stackoverflow.com/a/8389368/2539771
import urllib
URL = "https://images-na.ssl-images-amazon.com/images/I/714tx9QbaKL.SL1500.jpg"
urllib.urlretrieve(URL, "sample.png")
from PIL import Image
img = Image.open('/home/sks/sample.png')
img.show()

How do I read image data from a URL?

What I'm trying to do is fairly simple when we're dealing with a local file, but the problem comes when I try to do this with a remote URL.
Basically, I'm trying to create a PIL image object from a file pulled from a URL. Sure, I could always just fetch the URL and store it in a temp file, then open it into an image object, but that feels very inefficient.
Here's what I have:
Image.open(urlopen(url))
It flakes out complaining that seek() isn't available, so then I tried this:
Image.open(urlopen(url).read())
But that didn't work either. Is there a Better Way to do this, or is writing to a temporary file the accepted way of doing this sort of thing?
In Python3 the StringIO and cStringIO modules are gone.
In Python3 you should use:
from PIL import Image
import requests
from io import BytesIO
response = requests.get(url)
img = Image.open(BytesIO(response.content))
Using a StringIO
import urllib, cStringIO
file = cStringIO.StringIO(urllib.urlopen(URL).read())
img = Image.open(file)
The following works for Python 3:
from PIL import Image
import requests
im = Image.open(requests.get(url, stream=True).raw)
References:
https://github.com/python-pillow/Pillow/pull/1151
https://github.com/python-pillow/Pillow/blob/master/CHANGES.rst#280-2015-04-01
Using requests:
from PIL import Image
import requests
from StringIO import StringIO
response = requests.get(url)
img = Image.open(StringIO(response.content))
Python 3
from urllib.request import urlopen
from PIL import Image
img = Image.open(urlopen(url))
img
Jupyter Notebook and IPython
import IPython
url = 'https://newevolutiondesigns.com/images/freebies/colorful-background-14.jpg'
IPython.display.Image(url, width = 250)
Unlike other methods, this method also works in a for loop!
Use StringIO to turn the read string into a file-like object:
from StringIO import StringIO
from PIL import Image
import urllib
Image.open(StringIO(urllib.request.urlopen(url).read()))
For those doing some sklearn/numpy post processing (i.e. Deep learning) you can wrap the PIL object with np.array(). This might save you from having to Google it like I did:
from PIL import Image
import requests
import numpy as np
from StringIO import StringIO
response = requests.get(url)
img = np.array(Image.open(StringIO(response.content)))
The arguably recommended way to do image input/output these days is to use the dedicated package ImageIO. Image data can be read directly from a URL with one simple line of code:
from imageio import imread
image = imread('https://cdn.sstatic.net/Sites/stackoverflow/img/logo.png')
Many answers on this page predate the release of that package and therefore do not mention it. ImageIO started out as component of the Scikit-Image toolkit. It supports a number of scientific formats on top of the ones provided by the popular image-processing library PILlow. It wraps it all in a clean API solely focused on image input/output. In fact, SciPy removed its own image reader/writer in favor of ImageIO.
select the image in chrome, right click on it, click on Copy image address, paste it into a str variable (my_url) to read the image:
import shutil
import requests
my_url = 'https://www.washingtonian.com/wp-content/uploads/2017/06/6-30-17-goat-yoga-congressional-cemetery-1-994x559.jpg'
response = requests.get(my_url, stream=True)
with open('my_image.png', 'wb') as file:
shutil.copyfileobj(response.raw, file)
del response
open it;
from PIL import Image
img = Image.open('my_image.png')
img.show()
Manually wrapping in BytesIO is no longer needed since PIL >= 2.8.0. Just use Image.open(response.raw)
Adding on top of Vinícius's comment:
You should pass stream=True as noted https://requests.readthedocs.io/en/master/user/quickstart/#raw-response-content
So
img = Image.open(requests.get(url, stream=True).raw)
USE urllib.request.urlretrieve() AND PIL.Image.open() TO DOWNLOAD AND READ IMAGE DATA :
import requests
import urllib.request
import PIL
urllib.request.urlretrieve("https://i.imgur.com/ExdKOOz.png", "sample.png")
img = PIL.Image.open("sample.png")
img.show()
or Call requests.get(url) with url as the address of the object file to download via a GET request. Call io.BytesIO(obj) with obj as the content of the response to load the raw data as a bytes object. To load the image data, call PIL.Image.open(bytes_obj) with bytes_obj as the bytes object:
import io
response = requests.get("https://i.imgur.com/ExdKOOz.png")
image_bytes = io.BytesIO(response.content)
img = PIL.Image.open(image_bytes)
img.show()
from PIL import Image
import cv2
import numpy as np
import requests
image=Image.open(requests.get("https://previews.123rf.com/images/darrenwhi/darrenwhi1310/darrenwhi131000024/24022179-photo-of-many-cars-with-one-a-different-color.jpg", stream=True).raw)
#image =resize((420,250))
image_array=np.array(image)
image
To directly get image as numpy array without using PIL
import requests, io
import matplotlib.pyplot as plt
response = requests.get(url).content
img = plt.imread(io.BytesIO(response), format='JPG')
plt.imshow(img)

Categories

Resources