我尝试实现一个可以分配字符串的图像字段,它会自动从该URL获取图像。之后读它,它将保存本地副本的路径。因此,我继承了Django的ImageField
及其描述符类。
import uuid
import urllib.request
from django.core.files.base import ContentFile
from django.db import models
from django.db.models.fields.files import ImageFileDescriptor, ImageFieldFile
class UrlImageFileDescriptor(ImageFileDescriptor):
def __init__(self, field):
super().__init__(field)
def __get__(self, instance, owner=None):
# Get path to local copy on the server
try:
file = super().__get__(instance)
print('Get path to local copy', file.url)
return file.url if file else None
except:
return None
def __set__(self, instance, value):
# Set external URL to fetch new image from
print('Set image from URL', value)
# Validations
if not value:
return
value = value.strip()
if len(value) < 1:
return
if value == self.__get__(instance):
return
# Fetch and store image
try:
response = urllib.request.urlopen(value)
file = response.read()
name = str(uuid.uuid4()) + '.png'
content = ContentFile(file, name)
super().__set__(instance, content)
except:
pass
class UrlImageField(models.ImageField):
descriptor_class = UrlImageFileDescriptor
保存使用此字段的模型时,Django代码会引发错误'str' object has no attribute '_committed'
。这是Django 1.7c1的相关代码。它位于db / models / fields / files.py中。异常发生在if语句的行上。
def pre_save(self, model_instance, add):
"Returns field's value just before saving."
file = super(FileField, self).pre_save(model_instance, add)
if file and not file._committed:
# Commit the file to storage prior to saving the model
file.save(file.name, file, save=False)
return file
我不明白file
在这里是一个字符串。我唯一能想到的是描述符类__get__
返回字符串。但是,它使用__get__
调用其基类的ContentFile
,因此应该存储在模型的__dict__
中。谁可以给我解释一下这个?我怎样才能找到解决方法?
答案 0 :(得分:1)
问题是你需要返回FieldFile
,这样你才能访问它的属性,在django的源代码中你可以找到一个名为FileDescriptor的类是ImageFileDescriptor的父级,如果您查看名称下的FileDescriptor
类,您可以找到该类的文档,并说:
"""
The descriptor for the file attribute on the model instance. Returns a
FieldFile when accessed so you can do stuff like::
>>> from myapp.models import MyModel
>>> instance = MyModel.objects.get(pk=1)
>>> instance.file.size
Assigns a file object on assignment so you can do::
>>> with open('/tmp/hello.world', 'r') as f:
... instance.file = File(f)
"""
因此,您需要返回FieldFile
而不是String
,只需更改此回报即可。
return None or file
<强>更新强>:
我发现了你的问题,这段代码对我有用:
import uuid
import requests
from django.core.files.base import ContentFile
from django.db import models
from django.db.models.fields.files import ImageFileDescriptor, ImageFieldFile
class UrlImageFileDescriptor(ImageFileDescriptor):
def __init__(self, field):
super(UrlImageFileDescriptor, self).__init__(field)
def __set__(self, instance, value):
if not value:
return
if isinstance(value, str):
value = value.strip()
if len(value) < 1:
return
if value == self.__get__(instance):
return
# Fetch and store image
try:
response = requests.get(value, stream=True)
_file = ""
for chunk in response.iter_content():
_file+=chunk
headers = response.headers
if 'content_type' in headers:
content_type = "." + headers['content_type'].split('/')[1]
else:
content_type = "." + value.split('.')[-1]
name = str(uuid.uuid4()) + content_type
value = ContentFile(_file, name)
except Exception as e:
print e
pass
super(UrlImageFileDescriptor,self).__set__(instance, value)
class UrlImageField(models.ImageField):
descriptor_class = UrlImageFileDescriptor
class TryField(models.Model):
logo = UrlImageField(upload_to="victor")
custom_field = TryField.objects.create(logo="url_iof_image or File Instance") will work!!