[英]how to save pil cropped image to image field in django
我正在尝试将裁剪后的图像保存到模型中。 我收到以下错误:
回溯(最近一次调用):文件“/mypath/lib/python2.7/site-packages/django/core/handlers/base.py”,第 132 行,在 get_response 响应=wrapped_callback(request, *callback_args, ** callback_kwargs) 文件“/mypath/lib/python2.7/site-packages/django/contrib/auth/decorators.py”,第 22 行,在 _wrapped_view 中返回 view_func(request, *args, **kwargs) 文件“/mypath/ views.py”,第 236 行,在 player_edit player.save() 文件“/mypath/lib/python2.7/site-packages/django/db/models/base.py”,第 734 行,保存 force_update=force_update, update_fields=update_fields) 文件 "/mypath/lib/python2.7/site-packages/django/db/models/base.py", line 762, in save_base updated = self._save_table(raw, cls, force_insert, force_update, using , update_fields) 文件“/mypath/lib/python2.7/site-packages/django/db/models/base.py”,第 824 行,在 _save_table for f in non_pks] 文件“/mypath/lib/python2.7/ site-packages/django/db/models/fields/files.py", line 313, in pre_save if file and not file._ 已提交:文件“/mypath/lib/python2.7/site-packages/PIL/Image.py”,第 512 行,在getattr 中引发 AttributeError(name) AttributeError: _committed
我处理表单提交的视图如下所示:
if request.method == 'POST':
form = PlayerForm(request.POST, request.FILES, instance=current_player)
if form.is_valid():
temp_image = form.cleaned_data['profile_image2']
player = form.save()
cropped_image = cropper(temp_image, crop_coords)
player.profile_image = cropped_image
player.save()
return redirect('player')
裁剪功能如下所示:
from PIL import Image
import Image as pil
def cropper(original_image, crop_coords):
original_image = Image.open(original_image)
original_image.crop((0, 0, 165, 165))
original_image.save("img5.jpg")
return original_image
这是将裁剪后的图像保存到模型的正确过程吗? 如果是这样,为什么我会收到上述错误?
谢谢!
该函数应如下所示:
# The crop function looks like this:
from PIL import Image
from django.core.files.base import ContentFile
def cropper(original_image, crop_coords):
img_io = StringIO.StringIO()
original_image = Image.open(original_image)
cropped_img = original_image.crop((0, 0, 165, 165))
cropped_img.save(img_io, format='JPEG', quality=100)
img_content = ContentFile(img_io.getvalue(), 'img5.jpg')
return img_content
对于 Python 版本 >= 3.5
from io import BytesIO, StringIO()
img_io = StringIO() # or use BytesIO() depending on the type
其余的事情与@phourxx 的回答很好用
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