使用FileField在Django中保存原始文件名

Question

def generate_uuid_file_name(self, filename):
    self.original_filename = filename
    extension = filename.rsplit('.', 1)[1]
    newfilename = uuid.uuid4().__str__() + '.' + extension
    return self.directory() + newfilename

class FileUpload(models.Model):
    original_filename  = models.CharField(max_length=128)
    fileobj            = models.FileField(upload_to=generate_uuid_file_name)

上传时,

{"errors": {"original_filename": ["This field is required."]}, "success": false}

将blank = True,null = True添加到FileUpload.original_filename允许上载成功但不保存原始文件名.在Django 1.5上.根据这篇文章,这应该工作.

Answer 1

在视图中执行此操作(在null = True之后,blank = True再次成为模型的一部分):

file_object = UploadFileForm.save(commit=False)
file_object.original_filename = request.FILES['file'].name
file_object.save()

请注意,您需要根据您的上下文等相应地更改上述代码