Question

我正在编写一个 django 应用程序来上传带有表单的文件目录。

这是我使用的允许上传目录的表格：

class FileFieldForm(forms.Form):
    file_field = forms.FileField(widget=forms.ClearableFileInput(attrs=
        {'multiple': True, 'webkitdirectory': True, 'directory': True}))

这是原始的 post 负载：

------WebKitFormBoundaryPbO3HkrKGbBwgD3sd1
Content-Disposition: form-data; name="csrfmiddlewaretoken"

F575Bgl4U9dzgwePPeSW2ISZKk5c3CnRoqFasdasD0Hep6nD0LnAAObXbF92SUa96NbO2
------WebKitFormBoundaryPbO3HkrKGbBwgDsd31
Content-Disposition: form-data; name="file_field";
filename="MainDir/SubDir1/1.jpg"
Content-Type: image/jpeg


------WebKitFormBoundaryPbOasd3HkrKGbBwgD31
Content-Disposition: form-data; name="file_field";
filename="MainDir/SubDir2/2.jpg"
Content-Type: image/jpeg

这是处理表单的视图：

class FileFieldView(FormView):
    form_class = FileFieldForm
    template_name = 'upload.html'
    success_url = 'upload'

    def post(self, request, *args, **kwargs):
        form_class = self.get_form_class()
        form = self.get_form(form_class)
        files = request.FILES.getlist('file_field')
        if form.is_valid():
            for f in files:
                pprint("Name of file is " + f._get_name() + ' ' + f.field_name, sys.stderr)
                new_file = FileModel(file=f)
                new_file.save()
            return self.form_valid(form)
        else:
            return self.form_invalid(form)

问题是 django 中的文件对象名称没有子目录名称。我假设中间件处理请求之一是从文件名中解析和删除子目录名称。有什么方法可以获取包含目录和子目录名称的原始文件名？

Answer 1

我相信这就是 Django 的实现方式。请参考Django's Upload Handler doc.

它有默认的上传处理程序 MemoryFileUploadHandler 和 TemporaryFileUploadHandler。它们都使用 UploadedFile 来处理文件，并且它有一个函数 _set_name，它采用文件的基本名称。

甚至有评论说为什么它采用基本名称：

def _set_name(self, name):
    # Sanitize the file name so that it can't be dangerous.
    if name is not None:
        # Just use the basename of the file -- anything else is dangerous.
        name = os.path.basename(name)

        # File names longer than 255 characters can cause problems on older OSes.
        if len(name) > 255:
            name, ext = os.path.splitext(name)
            ext = ext[:255]
            name = name[:255 - len(ext)] + ext

    self._name = name

但我认为您可以编写自己的上传处理程序，它不采用基本名称并按照您的意愿运行。这里有一些关于如何编写 custom upload handler.

的信息

然后您需要在 FILE_UPLOAD_HANDLERS 设置中定义您的处理程序。

编辑 Custom Upload Handlers 使用 Django 3.1

Answer 2

扩展上一个答案，从目录上传中获取完整路径的一种方法是将文件路径（已清除）中的斜杠（\ 和 /）替换为连字符：

class CustomMemoryFileUploadHandler(MemoryFileUploadHandler):
    def new_file(self, *args, **kwargs):
        args = (args[0], args[1].replace('/', '-').replace('\', '-')) + args[2:]
        super(CustomMemoryFileUploadHandler, self).new_file(*args, **kwargs)

class CustomTemporaryFileUploadHandler(TemporaryFileUploadHandler):
    def new_file(self, *args, **kwargs):
        args = (args[0], args[1].replace('/', '-').replace('\', '-')) + args[2:]
        super(CustomTemporaryFileUploadHandler, self).new_file(*args, **kwargs)

@csrf_exempt
def my_view(request):
    # replace upload handlers. This depends on FILE_UPLOAD_HANDLERS setting. Below code handles the default in Django 1.10
    request.upload_handlers = [CustomMemoryFileUploadHandler(request), CustomTemporaryFileUploadHandler(request)]
    return _my_view(request)

@csrf_protect
def _my_view(request):
    # if the path of the uploaded file was "test/abc.jpg", here it will be "test-abc.jpg"
    blah = request.FILES[0].name

Answer 3

除了前面的答案之外，还有另一种可能对某人有用的方法。如果请求中只有 一个文件，您可以获得 multipart/form-data 负载中的原始文件名，而无需覆盖处理程序。

MemoryFileUploadHandler 和 TemporaryFileUploadHandler（默认使用，请参阅 Django's docs: Built-in upload handlers) are inherited from the FileUploadHandler class. Such objects have the file_name variable (see Django's code）。来自请求的文件之一的全名存储在这里（任何一个文件，我们不能提前说）。但是，如果您在请求中始终只有一个文件 - 就是这种方式。

因此视图将如下所示：

def your_view(request):
    file = request.FILES.get('file_field')
    full_file_name = request.upload_handlers[0].file_name # e.g. 'MainDir/SubDir1/1.jpg'

对于多文件上传，我们可以覆盖处理程序：

class NamedMemoryFileUploadHandler(MemoryFileUploadHandler):
    def file_complete(self, file_size):
        in_memory_file = super().file_complete(file_size)
        if in_memory_file is None:
            return
        return in_memory_file, self.file_name


class NamedTemporaryFileUploadHandler(TemporaryFileUploadHandler):
    def file_complete(self, file_size):
        temporary_file = super().file_complete(file_size)
        if temporary_file is None:
            return
        return temporary_file, self.file_name

@csrf_exempt
def upload_files(request):
    request.upload_handlers = [
        NamedMemoryFileUploadHandler(request),
        NamedTemporaryFileUploadHandler(request),
    ]
    return _upload_files(request)


@csrf_protect
def _upload_files(request):
    files = request.FILES.getlist("file") # list of tuples [(<file1>, "'MainDir/SubDir1/1.jpg'"), (<file2>, "'MainDir/SubDir2/2.jpg'")]
    for tmp_file, full_path in files:
        ...

Django目录上传获取子目录名

Django directory upload get sub-directory names

python

django

file-upload

django-forms

directory-upload