Django 动态 FileField upload_to

Question

我正在尝试为 FileField 模型制作动态上传路径。因此，当用户上传文件时，Django 将其存储到我的计算机 /media/(username)/(path_to_a_file)/(filename)。

例如 /media/Michael/Homeworks/Math/Week_1/questions.pdf 或 /media/Ernie/Fishing/Atlantic_ocean/Good_fishing_spots.txt

VIEWS
@login_required
def add_file(request, **kwargs):
if request.method == 'POST':
    form = AddFile(request.POST, request.FILES)
    if form.is_valid():
        post = form.save(commit=False)
        post.author = request.user

        post.parent = Directory.objects.get(directory_path=str(kwargs['directory_path']))
        post.file_path = str(kwargs['directory_path'])

        post.file_content = request.FILES['file_content'] <-- need to pass dynamic file_path here

        post.save()
        return redirect('/home/' + str(post.author))

MODELS
class File(models.Model):
    parent = models.ForeignKey(Directory, on_delete=models.CASCADE)
    author = models.ForeignKey(User, on_delete=models.CASCADE)
    file_name = models.CharField(max_length=100)
    file_path = models.CharField(max_length=900)
    file_content = models.FileField(upload_to='e.g. /username/PATH/PATH/..../')

FORMS
class AddFile(forms.ModelForm):
    class Meta:
        model = File
        fields = ['file_name', 'file_content']

我发现的是这个，但经过反复试验，我还没有找到方法。所以“上传/...”将是 post.file_path，它是动态的。

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)

Answer 1

你可以使用这样的东西（我在我的项目中使用过）：

import os
def get_upload_path(instance, filename):
    return os.path.join(
      "user_%d" % instance.owner.id, "car_%s" % instance.slug, filename)

现在：

photo = models.ImageField(upload_to=get_upload_path)

Answer 2

由于file_path是File模型上的一个属性，您是否可以不构建如下完整路径：

import os

def create_path(instance, filename):
    return os.path.join(
        instance.author.username,
        instance.file_path,
        filename
    )

然后从您的File模型中引用它：

class File(models.Model):
    ...
    file_content = models.FileField(upload_to=create_path)

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