We Keep Coding

How to work with uploaded file using Django

I have Optical Character Recognition (OCR) project. I am generating an API using Django Framework. The API should look like as below:
{
"id": 1,
"title": "PDF Title",
"input": "input.pdf",
"output": "output.pdf"
}
My models.py file as below:
from django.db import models
from .create_pdf_output import *
# Create your models here.
class Document(models.Model):
title = models.CharField(max_length=255)
pdf_input = models.FileField(upload_to='documents/inputs', max_length=200, blank=False)
pdf_output = models.FileField(upload_to='documents/outputs', max_length=200, blank=True)
def save(self):
self.pdf_output = create_pdf_output(self.pdf_input)
super(Document, self).save()
def __str__(self):
return self.title
In models.py file I call create_pdf_file.py which should perform OCR operations on pdf input
create_pdf_output.py
import shutil
import cv2
import numpy as np
import img2pdf
from pdf2image import convert_from_path
import os
def create_pdf_output(pdf_input):
pdf = str(pdf_input)
pdf_name = os.path.splitext(pdf)[0] + "_out.pdf"
pages = convert_from_path(pdf_input, 500)
if not os.path.exists('images'):
os.mkdir('images')
for ind, page in enumerate(pages):
page.save(f'images/out{ind}.jpg', 'JPEG')
for ind, img in enumerate(os.listdir('images')):
img_rgb = cv2.imread(os.path.join('images', img))
img_gray = cv2.cvtColor(img_rgb, cv2.COLOR_BGR2GRAY)
template = cv2.imread('templates.jpg', 0)
w, h = template.shape[::-1]
res = cv2.matchTemplate(img_gray, template, cv2.TM_CCOEFF_NORMED)
threshold = 0.8
loc = np.where(res >= threshold)
for pt in zip(*loc[::-1]):
cropped_image = img_rgb[pt[1]:pt[1]+h, pt[0]:pt[0]+ 4*h]
blurred = cv2.blur(cropped_image, (50,50))
img_rgb[pt[1]:pt[1]+h, pt[0]:pt[0]+ 4*h] = blurred
if not os.path.exists('results'):
os.mkdir('results')
cv2.imwrite(f'results/res{ind}.jpg', img_rgb)
dirname = 'images'
with open(pdf_name, 'wb') as f:
imgs = []
for fname in os.listdir(dirname):
if not fname.endswith('.jpg'):
continue
path = os.path.join(dirname, fname)
if os.path.isdir(path):
continue
imgs.append(path)
f.write(img2pdf.convert(imgs))
shutil.rmtree('images')
shutil.rmtree('results')
return pdf_name
However, since create_input_output function takes FieldFile as input it generates an error as below.
expected str, bytes or os.PathLike object, not FieldFile
My question is how can I work with input file, so I can generate output pdf file?

views.py
# pdf api
class OcrPDfDataApiView(APIView):
parser_class = [MultiPartParser,]
def post(self, request, format=None):
if 'data' not in request.data:
raise ParseError("Empty content")
filename = "whatever.pdf" # received file name
file_obj = request.data['data']
with default_storage.open('tmp/'+filename, 'wb+') as destination:
for chunk in file_obj.chunks():
destination.write(chunk)
dirName = os.path.dirname(__file__)
cwd = Path.cwd()
filename = os.path.join(cwd, 'media','tmp','whatever.pdf')
texts = print_pages(filename)
return Response(texts)
utils.py
import pytesseract
from PIL import Image
from PIL import ImageFilter
import requests
import docx2txt
import pdf2image
import PyPDF2
import pyttsx3
import os
import io
def print_pages(pdf_file):
images = pdf2image.convert_from_path(pdf_file)
for pages,img in enumerate(images):
text = pytesseract.image_to_string(img)
print(text)
return text

how to override django files?

I have a function that crops user images but I don't want the model to have 2 fields so I made a function that overrides the original file and I noticed that the function works well on normal files but when I add the function to the view new file is made but at the media directory not even the the specified folder so how can i override files by Django ?
models.py
# defining directory for every patient
def user_directory_path(instance, filename):
# file will be uploaded to MEDIA_ROOT/patient_<id>/<filename>
return 'patient_{0}/{1}'.format(instance.patient.id, filename)
class UploadedImages(models.Model):
patient = models.ForeignKey(Patient,on_delete=models.CASCADE,related_name='images')
pre_analysed = models.ImageField(upload_to = user_directory_path ,
verbose_name = 'Image')
upload_time = models.DateTimeField(default=timezone.now)
the cropping function:
import os
from PIL import Image
def crop(corrd, file ,path,pk):
image = Image.open(file) #To open the image as file.path won't open
path_1 , fn = os.path.split(path) #which is MEDIA_ROOT/and <filename>
patient_dir = 'patient_{}'.format(pk) #to get the user directory
path_ = path_1+patient_dir+fn #MEDIA_ROOT/patient_<id>/<filename>
cropped_image = image.crop(corrd)
resized_image = cropped_image.resize((384, 384), Image.ANTIALIAS)
resized_image.save(path_)
return path_
views.py
if form.is_valid():
image = form.save(commit=False)
x = float(request.POST.get('x'))
y = float(request.POST.get('y'))
w = float(request.POST.get('width'))
h = float(request.POST.get('height'))
print(x)
print(y)
print(w)
print(h)
crop((x,y,w+x,y+h),image.pre_analysed,image.pre_analysed.path)
image.patient = patient
messages.success(request,"Image added successfully!")
image.save()
forms.py
class ImageForm(ModelForm):
x = forms.FloatField(widget=forms.HiddenInput())
y = forms.FloatField(widget=forms.HiddenInput())
width = forms.FloatField(widget=forms.HiddenInput())
height = forms.FloatField(widget=forms.HiddenInput())
class Meta:
model = UploadedImages
fields = ('pre_analysed', 'x', 'y', 'width', 'height', )
so what do I need to do here?
thanks in advance.

The path you get from .path is relative to your MEDIA_ROOT folder, which may prevent your crop function from being able to open() the image file.
You can then make a new cropped file somewhere in your MEDIA_ROOT. Be sure to use os.makedirs() to create all the directories between here and the file. Also I don't see you saving the path returned from crop() back into the ImageField.

Testing an ImageField in DRF

I am using Django Imagekit in models.py:
from imagekit.models import ProcessedImageField
class AltroUser(models.Model):
first_name = models.CharField(_('first name'), max_length=30)
image = ProcessedImageField(upload_to='media/path',
default='user_default.jpg',
processors=[ResizeToFill(640, 640)],
format='JPEG',
options={'quality': 60})
serializers.py:
class UserRegistrationSerializer(Serializer):
first_name = serializers.CharField()
image = serializers.ImageField()
I am trying to test the image field. I have tried following methods:
def get_test_image():
try:
image = DjangoFile(open(os.path.join(django_settings.MEDIA_ROOT, 'user_default.jpg'),
mode='rb'))
return image
except (OSError, IOError) as e:
return None
def get_test_image1():
file = io.BytesIO()
image = Image.new('RGBA', size=(100, 100), color=(155, 0, 0))
image.save(file, 'png')
file.name = 'test.png'
file.seek(0)
return SimpleUploadedFile('abc.jpg', file.read())
def get_test_image2():
path = os.path.join(django_settings.MEDIA_ROOT, 'user_default.jpg')
file = File(open(path, 'r+b'))
return SimpleUploadedFile('abc.jpg', file.read())
I have tried calling the above three methods to set the value of the image key but none of them worked.
For get_test_image(), I get a response "The submitted file is empty."
For get_test_image1() and get_test_image2(), I get a response "The submitted file is empty." with an exception before in the data image field '_io.BytesIO' object has no attribute 'encoding'.
I don't understand what am I missing. Please help.

Here's an example of what I'm using to generate test image:
from StringIO import StringIO
from PIL import Image
from django.core.files import File
def get_image_file(name='test.png', ext='png', size=(50, 50), color=(256, 0, 0)):
file_obj = StringIO()
image = Image.new("RGBA", size=size, color=color)
image.save(file_obj, ext)
file_obj.seek(0)
return File(file_obj, name=name)

In Pillow==8.3.1, using StringIO() gave me this error:
TypeError: string argument expected, got 'bytes'.
I used BytesIO() instead and it worked like a charm
from io import BytesIO
def get_image_file(
self, name="test.png", ext="png", size=(50, 50), color=(256, 0, 0)
):
file_obj = BytesIO()
image = Image.new("RGBA", size=size, color=color)
image.save(file_obj, ext)
file_obj.seek(0)
return File(file_obj, name=name)

PIL to Django ImageField

I try to create an image from an url and save it in my django model. If the first part works fine, I do not know how to associate the generated file to my object.
This is my function to generate the image file:
def get_remote_image(image_url, merchant_product_path):
im = None
name = ''
r = requests.get(image_url, stream=True)
if r.status_code == 200:
name = urlparse(image_url).path.split('/')[-1]
full_path = os.path.join(settings.MEDIA_ROOT, merchant_product_path)
if not os.path.exists(full_path):
os.makedirs(full_path)
im = Image.open(r.raw)
if im.mode != "RGB":
im = im.convert("RGB")
im.thumbnail((500, 500), Image.ANTIALIAS)
im.save(full_path + name, 'JPEG')
return {'im': im, 'name': name}
And now, the part to associate this file to my object:
i = get_remote_image(row['pict'], m.get_products_media_path())
obj, created = ProductLine.objects.update_or_create(
...
...
...
)
if i['im'] is not None:
try:
obj.main_picture.save(
i['name'],
ContentFile(i['im']),
save=True)
except TypeError:
continue
This code works but unfortunately, mu pictures are created in the correct folder, objects are created/update but each one has no picture file :(
Can someone tell me what's wrong ?

I've finally found a solution:
def get_remote_image(image_url):
im = None
name = ''
r = requests.get(image_url, stream=True)
if r.status_code == 200:
name = urlparse(image_url).path.split('/')[-1]
i = Image.open(r.raw)
buffer = BytesIO()
if i.mode != "RGB":
i = i.convert("RGB")
i.thumbnail((500, 500), Image.ANTIALIAS)
i.save(buffer, format='JPEG')
im = InMemoryUploadedFile(
buffer,
None,
name,
'image/jpeg',
buffer.tell(),
None)
return {'im': im, 'name': name}
and then:
obj, created = ProductLine.objects.update_or_create(
...
...
...
)
i = get_remote_image(row['pict'])
obj.main_picture.save(
os.path.join(m.get_products_image_path(), i['name']),
i['im'],
save=True)
Hope this will help some other users in this situation.

With a model like :
class ProductLine(models.Model):
name = models.CharField(max_length=250, null=True)
image = models.ImageField(null=True)
You can directly link the picture on your computer using is path instead of his binary content.
obj, created = ProductLine.objects.update_or_create(...)
obj.image.name = "/path/to/the/file"
obj.save()

Django create thumbnail with PIL when i upload image

I use django-dropbox for the image storage on my website.
When i upload the image i want to automatically create an thumbnail and save it.
I tryed different methods but with no success
Any help will be apreciated
I get this exception error:
cannot identify image file
Here is the code:
class Product(models.Model):
image = models.ImageField(upload_to='images', storage=STORAGE, null=True, blank=True)
thumb = models.ImageField(upload_to='thumbs', storage=STORAGE, null=True, blank=True)
def __unicode__(self):
return "%s,%s" %(self.id,self.title)
def save(self):
# create a thumbnail
self.create_thumbnail()
super(Product, self).save()
def create_thumbnail(self):
# create a thumbnail
#self.create_thumbnail()
from PIL import Image
from cStringIO import StringIO
from django.core.files.uploadedfile import SimpleUploadedFile
import os
try:
DJANGO_TYPE = self.image.file.content_type
print DJANGO_TYPE
if DJANGO_TYPE == 'image/jpeg':
PIL_TYPE = 'jpeg'
FILE_EXTENSION = 'jpg'
elif DJANGO_TYPE == 'image/png':
PIL_TYPE = 'png'
FILE_EXTENSION = 'png'
else:
print error
print "Working1"
print StringIO(self.image.read())
im = Image.open(StringIO(self.image.read()))
size = 128, 128
im.thumbnail(size, Image.ANTIALIAS)
# Save the thumbnail
print "Working2"
temp_handle = StringIO()
im.save(temp_handle, PIL_TYPE)
temp_handle.seek(0)
# Save image to a SimpleUploadedFile which can be saved into
# ImageField
print "Working3"
suf = SimpleUploadedFile(os.path.split(self.image.name)[-1],
temp_handle.read(), content_type=DJANGO_TYPE)
# Save SimpleUploadedFile into image field
self.thumb.save('%s_thumbnail.%s'%(os.path.splitext(suf.name)[0],FILE_EXTENSION), suf, save=True)
print "Working4"
except Exception as e:
print e