54 lines
1.8 KiB
Python
54 lines
1.8 KiB
Python
import os
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from io import BytesIO
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from PIL import Image
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def rescale_image_jpeg(image_object, base_width=1020):
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img = Image.open(image_object)
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icc_profile = img.info.get('icc_profile') # remember color profile to not mess up colors
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width_percent = (base_width / float(img.size[0]))
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height = int((float(img.size[1]) * float(width_percent)))
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img = img.resize((base_width, height), Image.LANCZOS)
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img_bytes = BytesIO()
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img.save(img_bytes, 'JPEG', quality=90, optimize=True, icc_profile=icc_profile)
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return img_bytes
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def rescale_image_png(image_object, base_width=1020):
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image_object = Image.open(image_object)
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wpercent = (base_width / float(image_object.size[0]))
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hsize = int((float(image_object.size[1]) * float(wpercent)))
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img = image_object.resize((base_width, hsize), Image.LANCZOS)
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im_io = BytesIO()
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img.save(im_io, 'PNG', quality=90)
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return im_io
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def get_filetype(name):
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try:
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return os.path.splitext(name)[1]
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except Exception:
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return '.jpeg'
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# TODO this whole file needs proper documentation, refactoring, and testing
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# TODO also add env variable to define which images sizes should be compressed
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# filetype argument can not be optional, otherwise this function will treat all images as if they were a jpeg
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# Because it's no longer optional, no reason to return it
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def handle_image(request, image_object, filetype):
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try:
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Image.open(image_object).verify()
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except Exception:
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return None
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if (image_object.size / 1000) > 500: # if larger than 500 kb compress
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if filetype == '.jpeg' or filetype == '.jpg':
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return rescale_image_jpeg(image_object)
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if filetype == '.png':
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return rescale_image_png(image_object)
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return image_object
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