Как я могу использовать DrawImage ReportLab с URL -адресом изображения? ⇐ Python

[Anonymous]

Когда я попробую следующее: < /p>

from reportlab.lib.pagesizes import letter
from reportlab.pdfgen.canvas import Canvas
import urllib
import StringIO
import PIL.Image

image_file = urllib.urlopen('https://www.google.com/images/srpr/logo11w.png')
image_string = StringIO.StringIO(image_file.read())
logo = PIL.Image.open(image_string)

canvas = Canvas('output.pdf', pagesize=letter)
canvas.drawImage(logo, 10, 10)
canvas.showPage()
canvas.save()
< /code>

Я получаю эту ошибку: < /p>

Traceback (most recent call last):
File "imagefromurl.py", line 12, in
canvas.drawImage(logo, 10, 10)
File "/usr/lib/python2.7/dist-packages/reportlab/pdfgen/canvas.py", line 857, in drawImage
imgObj = pdfdoc.PDFImageXObject(name, image, mask=mask)
File "/usr/lib/python2.7/dist-packages/reportlab/pdfbase/pdfdoc.py", line 2090, in __init__
ext = string.lower(os.path.splitext(source)[1])
File "/usr/lib/python2.7/posixpath.py", line 96, in splitext
return genericpath._splitext(p, sep, altsep, extsep)
File "/usr/lib/python2.7/genericpath.py", line 91, in _splitext
sepIndex = p.rfind(sep)
File "/usr/lib/python2.7/dist-packages/PIL/Image.py", line 515, in __getattr__
raise AttributeError(name)
< /code>

reportlab - версия 2.5. < /p>

Подробнее здесь: https://stackoverflow.com/questions/261 ... -image-url