pillow OSError: cannot identify image file <_io.BytesIO object at 0x02345ED8>

Question

I'm trying to get a jpeg, gif, etc. from http requests but I'm getting an error. Here is a part of the code:

def printimg(url):

    http = httppoolmgr()

    file = http.request('GET',url).read()
    r_data = binascii.unhexlify(file)
    stream = io.BytesIO(r_data)
    img = Image.open(stream)
    #img = Image.open(file)

    return img

the error code return is :

  File "C:\Python34\lib\site-packages\pillow-2.5.3-py3.4-win32.egg\PIL\Image.py"
, line 2256, in open
OSError: cannot identify image file <_io.BytesIO object at 0x02345ED8>

Has anyone already met this kind of problem?

Not sure what your intent is here but you could use urllib to retrieve the image and then open as Image object and return. Check: https://docs.python.org/2/library/urllib.html#urllib.urlretrieve — user3885927, Oct 01 '14 at 21:59
Ok I think also but urllib3 bring new functionalities and I would like to use only this librairy if possible — JroS, Oct 02 '14 at 16:11
Duplicate of http://stackoverflow.com/questions/7391945/how-do-i-read-image-data-from-a-url-in-python? Image.open() expects a filename — Hugo, Oct 04 '14 at 06:02

score -1 · Answer 1 · edited Dec 15 '15 at 07:14

I replace the retrieve part of the image by the following code

response = requests.get(url)
img = Image.open(io.BytesIO(response.content))
img.save("picture/%s.png" % row)
self.foto = PhotoImage(file="picture/%s.png" % row)

Label(self.frame, image=self.foto, name=str(row)).grid(row=row, column=0, sticky=W)

Now there is no error at all about opening image or something else now I m getting an other problem with the following code:

def populate(self):

    http = httppoolmgr()
    array = xmltohash(getrack(http,'618cd2a4a2a1740a9f46e4f367ef88f3'))

    for row in range(len(array)):
        url = str((array[row]),"utf-8").split("$#$")[3]
        title = str((array[row]),"utf-8").split("$#$")[1]
        response = requests.get(url)
        img = Image.open(io.BytesIO(response.content))
        img.save("picture/%s.png" % row)
        self.foto = PhotoImage(file="picture/%s.png" % row)

        Label(self.frame, image=self.foto, name=str(row)).grid(row=row, column=0, sticky=W)
        t=str((array[row]),"utf-8").split("$#$")[1]
        Label(self.frame, text=t).grid(row=row, column=1, sticky=W)

Only the last image is displayed but all the text are displayed correctly in the frame. If someone can help please I'm a little bit stuck

Regards,

JRoS

Please don't ask a second question in your answer. You can post such things as new questions if need be, and link to the previous question if warranted for context. — CrazyChucky, Nov 12 '21 at 10:08

score -1 · Answer 2 · answered Oct 06 '14 at 19:37

ok this the solution:

def populate(self):

    http = httppoolmgr()
    array = xmltohash(getrack(http,'618cd2a4a2a1740a9f46e4f367ef88f3'))

    for row in range(len(array)):
        url = str((array[row]),"utf-8").split("$#$")[3]
        title = str((array[row]),"utf-8").split("$#$")[1]
        response = requests.get(url)
        img = Image.open(io.BytesIO(response.content))
        img.save("picture/%s.png" % row)
        self.foto = PhotoImage(file="picture/%s.png" % row)

        pic=Label(self.frame, image=self.foto, name=str(row))
        pic.image = self.foto
        pic.grid(row=row, column=0, sticky=W)
        t=str((array[row]),"utf-8").split("$#$")[1]
        Label(self.frame, text=t).grid(row=row, column=1, sticky=W)

for my point of view there is no sense why the previous code was not working but I guess python or tkinter doesn't like to cast many options woinw no I know :D

pillow OSError: cannot identify image file <_io.BytesIO object at 0x02345ED8>

2 Answers2