我意识到这是一项微不足道的任务,这个问题已被多次回答,但我无法得到它。有没有办法在将图像保存到磁盘之前调整大小并裁剪图像?我发现的所有解决方案都倾向于存储图像,调整大小,然后再次存储。我可以这样做吗?
# extending form's save() method
def save(self):
import Image as pil
# if avatar is uploaded, we need to scale it
if self.files['avatar']:
img = pil.open(self.files['avatar'])
img.thumbnail((150, 150), pil.ANTIALIAS)
# ???
# self.files['avatar'] is InMemoryUpladedFile
# how do I replace self.files['avatar'] with my new scaled image here?
# ???
super(UserForm, self).save()
答案 0 :(得分:6)
我能够弄清楚这一点。您只需将修改后的文件保存为StringIO
,然后从中创建新的InMemoryUploadedFile
即可。这是完整的解决方案:
def save(self):
import Image as pil
import StringIO, time, os.path
from django.core.files.uploadedfile import InMemoryUploadedFile
# if avatar is uploaded, we need to scale it
if self.files['avatar']:
# opening file as PIL instance
img = pil.open(self.files['avatar'])
# modifying it
img.thumbnail((150, 150), pil.ANTIALIAS)
# saving it to memory
thumb_io = StringIO.StringIO()
img.save(thumb_io, self.files['avatar'].content_type.split('/')[-1].upper())
# generating name for the new file
new_file_name = str(self.instance.id) +'_avatar_' +\
str(int(time.time())) + \
os.path.splitext(self.instance.avatar.name)[1]
# creating new InMemoryUploadedFile() based on the modified file
file = InMemoryUploadedFile(thumb_io,
u"avatar", # important to specify field name here
new_file_name,
self.files['avatar'].content_type,
thumb_io.len,
None)
# and finally, replacing original InMemoryUploadedFile() with modified one
self.instance.avatar = file
# now, when form will be saved, it will use the modified file, instead of the original
super(UserForm, self).save()
答案 1 :(得分:0)
我不熟悉PIL,但正如我从文档中看到的那样,您可以将文件对象作为“文件”参数传递给“打开”函数。
Django request.FILES
存储UploadedFile对象 - 围绕上传文件的简单包装(存储在内存或临时文件中),它支持read,seek和tell操作,因此可以直接传递给PIL“open”功能