Opening image from http in python 3

Question

I know there are a thousand questions like that out there but none of the provided solutions do actually work. I am using pythong 3.4. I would like to directly open the url as an image, not store it first on the disk.

Basically the code comes down to this

from PIL import Image <br />
import urllib.request <br />
... <br />
opener = urllib.request.build_opener() <br />
file = opener.open("http://pixel.quantserve.com/pixel") <br />
img = Image.open(file.read()) <br />
width, height = img.size

This yields for me an error. I also tried it without the .read() and with
file = urllib.request.urlopen("...")
both with and without the .read()

Basically I am lost. The only thing I can do is change the error that python is throwing at me. Thanks for any help!
Error messages with the above mentioned version:
TypeError: embedded NUL character
Without the read()
io.UnsupportedOperation: seek

With "urllib.request.urlopen("...") without .read()
io.UnsupportedOperation: seek

with .read()
TypeError: embedded NUL character

First of all, format your code correctly, What is the error that python throws at you? — no_name
– no_name, Commented Feb 7, 2015 at 14:38

Paul · Accepted Answer · 2015-02-07 15:45:35Z

4

To avoid using tkinter and to avoid writing the file locally, use a buffer:

from PIL import Image
import urllib
from io import BytesIO

f = urllib.urlopen("http://pixel.quantserve.com/pixel")
b = BytesIO(f.read())
i = Image.open(b)

i # <PIL.TgaImagePlugin.TgaImageFile image mode=P size=512x17410 at 0x7FBA4A3712D8>

This is using 2.7, which doesn't have urllib.request, so modify this as required.

Also, be aware that this image appears to be malformed/invalid

edited Feb 7, 2015 at 15:45

answered Feb 7, 2015 at 15:38

Paul

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Paul Over a year ago

@notsoimportant The best way to say this is to mark the answer as correct!

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