4

可能重复:
Django-tastypie:POST 中文件上传的任何示例?

我目前像这样向我的 API 发出 cURL POST 请求

curl --dump-header - -H "Content-Type: application/json" -X POST --data '{"username":"theusername", "api_key":"anapikey", "video_title":"a title", "video_description":"the description"}' http://localhost:8000/api/v1/video/ 

但现在我需要能够将视频文件添加到上传中。关于使用 Tastypie 上传文件的问题,我已经四处寻找了几个小时,但我没有提出一个可靠的回应。我需要添加 Base64 编码吗?如果有怎么办?使用 POST 请求上传文件后,如何访问该文件?只是正常的 request.FILES 动作?我不希望将文件保存到数据库,只是获取文件的路径。

#Models.py
class Video(models.Model):
    video_uploader = models.ForeignKey(User)
    video_path = models.CharField(max_length=128)
    video_views = models.IntegerField(default=0)
    upload_date = models.DateTimeField(auto_now_add=True)
    video_description = models.CharField(max_length=860)
    video_title = models.SlugField()

我对如何为 Tastypie 实现文件上传系统感到非常困惑,因此非常感谢任何帮助。谢谢!

4

1 回答 1

13

这是MultiPart通过django-tastypie.

模型.py

class Video(models.Model):
    video_uploader = models.ForeignKey(User)
    video = models.FileField(_('Video'), upload_to='path_to_folder/') # save file to server
    video_views = models.IntegerField(default=0)
    upload_date = models.DateTimeField(auto_now_add=True)
    video_description = models.CharField(max_length=860)
    video_title = models.SlugField()

api.py

class MultipartResource(object):
    def deserialize(self, request, data, format=None):
        if not format:
            format = request.META.get('CONTENT_TYPE', 'application/json')
        if format == 'application/x-www-form-urlencoded':
            return request.POST
        if format.startswith('multipart'):
            data = request.POST.copy()
            data.update(request.FILES)
            return data
        return super(MultipartResource, self).deserialize(request, data, format)

class VideoResource(MultipartResource, ModelResource):
   """
   Inherit this Resource class to `MultipartResource` Class
   """
   # Assuming you know what to write here 
   ...

然后通过CURL

curl -H "Authorization: ApiKey username:api_key" -F "video=/path_to_video/video.mp3" -F "video_title=video title" http://localhost:8000/api/v1/video/ -v
于 2013-01-03T07:39:41.447 回答