python - 在验证和保存表单之前，如何从 ImageField 打开图像？

Question

我想根据 ImageField 的值获取 CharField 的值。我的表单和 View 定义为:

#Form
class GpsImForm(forms.Form):
    image = forms.ImageField(required=True)
    gps_data = forms.CharField(required=True)

#View
def gpsim_gen_view(request):
    if request.method == 'POST':
        form = GpsImForm(request.POST, request.FILES)
        if 'image' in request.FILES:
            im = request.FILES['image']
            i = Image.open(im)
            ... # functions to extract exif data from i
            request.POST.update({ 'gps_data': ...}) # set gps_data based on exif data from i
        if form.is_valid():
            obj = form.save()
            return ... #returns the gpsim
    else:
        form = GpsImForm()
    return direct_to_template(request, 'gpsim_generate.html', {'form': form,})

gps_data 已更新，但是，一旦我使用 Image.open()，我就会收到以下错误消息:

Upload a valid image. The file you uploaded was either not an image or a corrupted image.

如果我注释有关i的行并将gps_data修改为任何内容，则表单(带有图像)将被保存而不会出现任何错误...

# i = Image.open(im)
# ...
# functions to extract exif data from i
request.POST.update({ 'gps_data': 'some text'}) # set gps_data to 'test'

Answer 1

首先，确保您的表单具有 enctype 标记

<form enctype="multipart/form-data" ... >    

尝试将img(所有 block )写入磁盘:

import Image
from random import choice
from django.conf import settings

random_file_name = getattr(settings, 'FILE_UPLOAD_TEMP_DIR', '/tmp')
random_file_name +=  '/' + ''.join([choice('abcdefg') for i in range(12)]) + '.jpg'

destination = open(random_file_name, 'wb+')
for chunk in request.FILES['image'].chunks():
    destination.write(chunk)
destination.close()

然后，您可以从磁盘打开它:

image = Image.open(random_file_name)
if image.mode != "RGB":
    image = image.convert("RGB")

...

此网址可能对您有帮助:

https://docs.djangoproject.com/en/dev/topics/http/file-uploads/