How to parse data-uri in python?

Question

HTML image elements have this simplified format:

<img src='something'>

That something can be data-uri, for example:

data:image/png;base64,iVBORw0KGg...

Is there a standard way of parsing this with python, so that I get content_type and base64 data separated, or should I create my own parser for this?

Solution 1

Split the data URI on the comma to get the base64 encoded data without the header. Call base64.b64decode to decode that to bytes. Last, write the bytes to a file.

from base64 import b64decode

data_uri = "data:image/png;base64,iVBORw0KGg..."

# Python 2 and <Python 3.4
header, encoded = data_uri.split(",", 1)
data = b64decode(encoded)

# Python 3.4+
# from urllib import request
# with request.urlopen(data_uri) as response:
#     data = response.read()

with open("image.png", "wb") as f:
    f.write(data)

Solution 2

Python since 3.4 has support for data-uri, under the hood using urllib.request.DataHandler.

from urllib.request import urlopen

with urlopen(data_uri) as response:
    data = response.read()

Solution 3

w3lib (a library used by Scrapy) has a function to parse data uris:

>>> from w3lib.url import parse_data_uri
>>> parse_data_uri('data:image/png;base64,iVBORw0KGg==')
ParseDataURIResult(media_type='image/png', media_type_parameters={}, data=b'\x89PNG\r\n\x1a')

Solution 4

This may help:

import re
from lxml import html

BASE_NAME = "image_"

source_code = """<img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAUA
AAAFCAYAAACNbyblAAAAHElEQVQI12P4//8/w38GIAXDIBKE0DHxgljNBAAO
9TXL0Y4OHwAAAABJRU5ErkJggg==" alt="Red dot" />
<img src="data:image/gif;base64,R0lGODlhAQABAIAAAAUEBAAAACwAAAAAAQABAAACAkQBADs=" alt="Black dot" />"""

tree = html.fromstring(source_code)

for i,image in enumerate(tree.xpath('//img[contains(@src, "data:image")]/@src')):
    image_type, image_content = image.split(',', 1)
    image_type = re.findall('data:image\/(\w+);base64', image_type)[0]
    with open("{}{}.{}".format(BASE_NAME, i, image_type), "wb") as f:
        f.write(image_content.decode('base64'))
    print "[*] '{}' image found with content: {}\n".format(image_type, image_content)

Output:

[*] 'png' image found with content: iVBORw0KGgoAAAANSUhEUgAAAAUA
AAAFCAYAAACNbyblAAAAHElEQVQI12P4//8/w38GIAXDIBKE0DHxgljNBAAO
9TXL0Y4OHwAAAABJRU5ErkJggg==

[*] 'gif' image found with content: R0lGODlhAQABAIAAAAUEBAAAACwAAAAAAQABAAACAkQBADs=

It will save every base64 image within <img> tags, with their respective file extension:

Prefixed by BASE_NAME + auto-increment digit(s) provided by enumerate + image_extension

Solution 5

Correcting JRodDynamite's post:

from base64 import decodestring

png_arr= "data:image/png;base64,iVBORw0KGg..."
png_arr = png_arr.split(",")
png_arr = png_arr[1]

fh = open("imageToSave.png", "wb")
fh.write(decodestring(png_arr))
fh.close()

View more solutions

Comments

[Andrey Belyak]

the prettiest solution imho: short and produces well-structured result

[Darkyen]

Just splitting on first comma is not necessarily correct, the MIME may contain comma as well, for example: data:video/webm; codecs=\"vp8, opus\";base64,GkXfowEAAAAAAAAfQoaBAUL3g...

[Darkyen]

And quotes won't help, because this is also possible: data:video/webm;codecs=vp8,opus;base64,GkXfo59...