How to send image generated by PIL to browser?

Question

I'm using flask for my application. I'd like to send an image (dynamically generated by PIL) to client without saving on disk.

Any idea how to do this ?

Flask doesn't seem to have solid support for streaming binary data that you can't generate with a Python generator. You'll probably have to buffer the image in memory and sent that. — millimoose, Oct 25 '11 at 00:46

score 184 · Answer 1 · edited May 29 '23 at 13:04

184

Here's a version without any temp files and the like (see here):

from io import StringIO

def serve_pil_image(pil_img):
    img_io = StringIO()
    pil_img.save(img_io, 'JPEG', quality=70)
    img_io.seek(0)
    return send_file(img_io, mimetype='image/jpeg')

To use in your code simply do

@app.route('some/route/')
def serve_img():
    img = Image.new('RGB', ...)
    return serve_pil_image(img)

edited May 29 '23 at 13:04

snoob dogg

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answered Apr 16 '12 at 08:11

Mr. Mr.

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Means you have to be able to hold the whole image in memory at once though, right? Might be an issue with huge images or other types of downloads. – Eli Aug 08 '13 at 03:42
@Eli for those cases it seems more sensible to use a temporary file. Anyway, huge dynamically generated files aren't common. – erickrf May 18 '14 at 21:35
This is a great solution for generating files in memory – dalethestirling Jul 08 '14 at 12:49
5

How do I insert it into the template I am returning? – scottydelta Nov 16 '14 at 11:06
How would one add headers to this? Since we are not using make_response? – mgmonteleone Jan 14 '15 at 22:53
1

Use on the side of caution. Some wsgi implementations require a `fileno()` method for sendfile, and StringIO objects has no `fileno()` method. – Kenneth Nov 27 '15 at 02:36
1

Assuming you're doing `import StringIO` you'll need to change a line: `img_io = StringIO.StringIO()` – Merkidemis Nov 09 '17 at 16:38
For python3 it's io.BytesIO (not ByteIO) – Johan Gyllenspetz May 24 '19 at 19:12
2

How would I now reference the file in the `src` attribute of an `img` tag – Riley Fitzpatrick Oct 02 '19 at 15:26
2

for those asking how to reference the file in the `src` attribute of an `img` tag, simply use an endpoint that will serve the images `@app.route('/docs/')` and then in the HTML use `src="/docs/some_img.jpg" – Green Jan 21 '20 at 12:01

score 36 · Answer 2 · answered Aug 23 '18 at 13:16

36

Mr. Mr. did an excellent job indeed. I had to use BytesIO() instead of StringIO().

def serve_pil_image(pil_img):
    img_io = BytesIO()
    pil_img.save(img_io, 'JPEG', quality=70)
    img_io.seek(0)
    return send_file(img_io, mimetype='image/jpeg')

answered Aug 23 '18 at 13:16

Dan Erez

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I'm having an issue with the fact that BytesIO is not a string and `send_file` requires a string for the path. How can I get the path from the BytesIO object? – Riley Fitzpatrick Oct 02 '19 at 15:23
I think bytes come out as a 8 bit object that can be interpreted as a string. Anyways - I'm not really sure why it's not working for you, try StringIO() maybe? – Dan Erez Oct 02 '19 at 16:23
1

It actually was working fine, I just wasn't referencing it correctly in my html. Now using `url_for()` and it is working fine. – Riley Fitzpatrick Oct 02 '19 at 19:42
I have it setup like that, but when I run my application inside docker i get `RuntimeError: Attempted implicit sequence conversion but the response object is in direct passthrough mode.` Any ideas where to set it? – Saintan Sep 22 '20 at 09:08

score 20 · Accepted Answer · edited May 23 '17 at 11:54

20

First, you can save the image to a tempfile and remove the local file (if you have one):

from tempfile import NamedTemporaryFile
from shutil import copyfileobj
from os import remove

tempFileObj = NamedTemporaryFile(mode='w+b',suffix='jpg')
pilImage = open('/tmp/myfile.jpg','rb')
copyfileobj(pilImage,tempFileObj)
pilImage.close()
remove('/tmp/myfile.jpg')
tempFileObj.seek(0,0)

Second, set the temp file to the response (as per this stackoverflow question):

from flask import send_file

@app.route('/path')
def view_method():
    response = send_file(tempFileObj, as_attachment=True, attachment_filename='myfile.jpg')
    return response

edited May 23 '17 at 11:54

Community

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answered Oct 25 '11 at 05:39

Adam Morris

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Doesn't work anymore: `TypeError: 'file' object is not callable` – letmaik Apr 24 '14 at 06:50
Which is the dynamically created file over here? – user1953366 Nov 05 '17 at 22:00
@user1953366 - in this example, the file being created is myfile.jpg, but I would seriously check out mr-mr's answer below, as it is much more efficient without the need to create a temporary file. – Adam Morris Nov 06 '17 at 14:57
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temp files are also written to disk? how is this the accepted response? – Dan Erez Aug 22 '18 at 12:37

score 7 · Answer 4 · edited May 08 '17 at 16:01

7

It turns out that flask provides a solution (rtm to myself!):

from flask import abort, send_file
try:
    return send_file(image_file)
except:
    abort(404)

edited May 08 '17 at 16:01

Funk Forty Niner

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answered Dec 08 '11 at 18:50

cybertoast

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score 6 · Answer 5 · answered Nov 16 '11 at 00:07

I was also struggling in the same situation. Finally, I have found its solution using a WSGI application, which is an acceptable object for "make_response" as its argument.

from Flask import make_response

@app.route('/some/url/to/photo')
def local_photo():
    print('executing local_photo...')
    with open('test.jpg', 'rb') as image_file:
        def wsgi_app(environ, start_response):
            start_response('200 OK', [('Content-type', 'image/jpeg')])
            return image_file.read()
        return make_response(wsgi_app)

Please replace "opening image" operations with appropriate PIL operations.

score 1 · Answer 6 · answered Feb 17 '21 at 07:50

1

        return send_file(fileName, mimetype='image/png')

answered Feb 17 '21 at 07:50

Gank

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How to send image generated by PIL to browser?

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