how to compress the image before uploading to s3 in django?

问题

I am working on an application where a user can upload an image. I want to reduce the size of the image in 200-500kb. This is my models.py file

class Report_item(models.Model):
    owner = models.ForeignKey(settings.AUTH_USER_MODEL)
    title = models.CharField(max_length=255, help_text='*Title for the post e.g. item identity')
    image = models.ImageField(default="add Item image",
                          upload_to=get_uplaod_file_name)


    def __str__(self):
        return self.title + "      " + str(self.publish)

And this is my views.py file

class ReportCreate(generic.CreateView):
model = Report_item
fields = ['title','image']

def get_form(self, form_class=None):
    if form_class is None:
        form_class = self.get_form_class()
    form = super(ReportCreate, self).get_form(form_class)
    form.fields['title'].widget = TextInput(
        attrs={'placeholder': '*Enter UID e.g. CBSE Marksheet Roll nunber 0506***'})
    return form

def form_valid(self, form):
    self.object = form.save(commit=False)
    self.object.owner = self.request.user
    self.object.save()
    return FormMixin.form_valid(self, form)

I am using Django 1.11 and S3 storage. Kindly help me to compress the image before uploading to s3.

回答1:

So we need to define a save method in models in order to compress the image before save. Following code help me what I want to achieve for my problem.

class Report_item(models.Model):
    owner = models.ForeignKey(settings.AUTH_USER_MODEL)
    title = models.CharField(max_length=255, help_text='*Title for the post e.g. item identity')
    
    image = models.ImageField(default="add Item image",
                              upload_to=get_uplaod_file_name)

    def save(self):
        # Opening the uploaded image
        im = Image.open(self.image)

        output = BytesIO()

        # Resize/modify the image
        im = im.resize((100, 100))

        # after modifications, save it to the output
        im.save(output, format='JPEG', quality=90)
        output.seek(0)

        # change the imagefield value to be the newley modifed image value
        self.image = InMemoryUploadedFile(output, 'ImageField', "%s.jpg" % self.image.name.split('.')[0], 'image/jpeg',
                                        sys.getsizeof(output), None)

        super(Report_item, self).save()

WIth the help of this 5 Mb image compress to 4 kb approx.

回答2:

Suppose you want to upload an image after resizing it. You can use below code, just pass the image object and you will get resized image in return.

def GetThumbnail(f):
    try:
        name = str(f).split('.')[0]
        image = Image.open(f)
        image.thumbnail((400, 400), Image.ANTIALIAS)
        thumbnail = BytesIO()
        # Default quality is quality=75
        image.save(thumbnail, format='JPEG', quality=50)
        thumbnail.seek(0)
        newImage = InMemoryUploadedFile(thumbnail,
                                   None,
                                   name + ".jpg",
                                   'image/jpeg',
                                   thumbnail.tell(),
                                   None)
        return newImage
    except Exception as e:
        return e

来源：https://stackoverflow.com/questions/52183975/how-to-compress-the-image-before-uploading-to-s3-in-django