我正在尝试上传文件并在媒体文件夹中组织目录结构。具体来说,我希望上传根据模型中的一个值创建子文件夹。我面临的问题是在视图中我向实例添加信息(在我的示例代码中,这是相关的profile
)。我想将这些信息用于子文件夹,但在我的模型中不存在,直到上传后的保存...
将信息导入Upload
以便创建子文件夹的适当方法是什么?
由于
型号:
class Upload(models.Model):
file = models.FileField(upload_to="upload/")
profile = models.ForeignKey(Profile, blank=True, null=True)
def get_upload_to(self, field_attname):
return 'upload/%d' % self.profile
查看:
def profile(request, profile_slug):
profile = Profile.objects.get(slug=profile_slug)
context_dict['profile'] = profile
if request.method=="POST":
for file in request.FILES.getlist('file'):
upload = UploadForm(request.POST, request.FILES)
if upload.is_valid():
newupload = upload.save(commit=False)
newupload.profile = profile
newupload.save()
else:
pass
upload=UploadForm()
context_dict['form'] = upload
return render(request, 'app/profile.html', context_dict)
解决方案,感谢xyres:
型号:
def get_upload_to(instance, filename):
return 'upload/%s/%s' % (instance.profile, filename)
class Upload(models.Model):
file = models.FileField(upload_to=get_upload_to)
profile = models.ForeignKey(Profile, blank=True, null=True)
查看:
def profile(request, profile_slug):
profile = Profile.objects.get(slug=profile_slug)
context_dict['profile'] = profile
if request.method=="POST":
upload = UploadForm(request.POST, request.FILES)
if upload.is_valid():
for f in request.FILES.getlist('file'):
Upload.objects.create(file=f, profile=profile)
return redirect(reverse('profile')
else:
pass
return render(request, 'app/profile.html', context_dict)
答案 0 :(得分:2)
将返回上传路径的函数需要2个参数,即:instance
和filename
。您需要在此类之外定义此函数,并将此函数提供给upload_to
选项。
这就是你需要重写代码的方法:
def get_upload_to(instance, filename):
return 'upload/%d/%s' % (instance.profile, filename)
class Upload(models.Model):
file = models.FileField(upload_to=get_upload_to)
profile = models.ForeignKey(Profile, blank=True, null=True)
修改强>
如果您想知道为什么无法在课堂内定义get_upload_to
,以下是您最有可能获得的错误:
ValueError:在categories.models中找不到函数get_upload_to。
请注意,由于Python 2的限制,您无法序列化未绑定的方法函数(例如,在同一个类体中声明和使用的方法)。请将该功能移动到主模块主体中以使用迁移。