59

我有一个 Django 应用程序，其视图接受要上传的文件。使用 Django REST 框架，我将 APIView 子类化并实现 post() 方法，如下所示：

class FileUpload(APIView):
    permission_classes = (IsAuthenticated,)

    def post(self, request, *args, **kwargs):
        try:
            image = request.FILES['image']
            # Image processing here.
            return Response(status=status.HTTP_201_CREATED)
        except KeyError:
            return Response(status=status.HTTP_400_BAD_REQUEST, data={'detail' : 'Expected image.'})

现在我正在尝试编写几个单元测试以确保需要身份验证并且实际处理了上传的文件。

class TestFileUpload(APITestCase):
    def test_that_authentication_is_required(self):
        self.assertEqual(self.client.post('my_url').status_code, status.HTTP_401_UNAUTHORIZED)

    def test_file_is_accepted(self):
        self.client.force_authenticate(self.user)
        image = Image.new('RGB', (100, 100))
        tmp_file = tempfile.NamedTemporaryFile(suffix='.jpg')
        image.save(tmp_file)
        with open(tmp_file.name, 'rb') as data:
            response = self.client.post('my_url', {'image': data}, format='multipart')
            self.assertEqual(status.HTTP_201_CREATED, response.status_code)

但是当 REST 框架尝试对请求进行编码时，这会失败

Traceback (most recent call last):
  File "/home/vagrant/.virtualenvs/myapp/lib/python3.3/site-packages/django/utils/encoding.py", line 104, in force_text
    s = six.text_type(s, encoding, errors)
UnicodeDecodeError: 'utf-8' codec can't decode byte 0xff in position 118: invalid start byte

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "/home/vagrant/webapp/myproject/myapp/tests.py", line 31, in test_that_jpeg_image_is_accepted
    response = self.client.post('my_url', { 'image': data}, format='multipart')
  File "/home/vagrant/.virtualenvs/myapp/lib/python3.3/site-    packages/rest_framework/test.py", line 76, in post
    return self.generic('POST', path, data, content_type, **extra)
  File "/home/vagrant/.virtualenvs/myapp/lib/python3.3/site-packages/rest_framework/compat.py", line 470, in generic
    data = force_bytes_or_smart_bytes(data, settings.DEFAULT_CHARSET)
  File "/home/vagrant/.virtualenvs/myapp/lib/python3.3/site-packages/django/utils/encoding.py", line 73, in smart_text
    return force_text(s, encoding, strings_only, errors)
  File "/home/vagrant/.virtualenvs/myapp/lib/python3.3/site-packages/django/utils/encoding.py", line 116, in force_text
    raise DjangoUnicodeDecodeError(s, *e.args)
django.utils.encoding.DjangoUnicodeDecodeError: 'utf-8' codec can't decode byte 0xff in position 118: invalid start byte. You passed in b'--BoUnDaRyStRiNg\r\nContent-Disposition: form-data; name="image"; filename="tmpyz2wac.jpg"\r\nContent-Type: image/jpeg\r\n\r\n\xff\xd8\xff[binary data omitted]' (<class 'bytes'>)

如何让测试客户端发送数据而不尝试将其解码为 UTF-8？

4

5 回答 5

50

测试文件上传时，您应该将流对象传递给请求，而不是数据。

@arocks在评论中指出了这一点

通过 { 'image': file} 代替

但这并没有完全解释为什么需要它（也与问题不符）。对于这个特定的问题，你应该做

from PIL import Image

class TestFileUpload(APITestCase):

    def test_file_is_accepted(self):
        self.client.force_authenticate(self.user)

        image = Image.new('RGB', (100, 100))

        tmp_file = tempfile.NamedTemporaryFile(suffix='.jpg')
        image.save(tmp_file)
        tmp_file.seek(0)

        response = self.client.post('my_url', {'image': tmp_file}, format='multipart')

       self.assertEqual(status.HTTP_201_CREATED, response.status_code)

这将匹配标准的 Django 请求，其中文件作为流对象传入，由 Django REST Framework 处理。当您只传入文件数据时，Django 和 Django REST Framework 将其解释为字符串，这会导致问题，因为它需要一个流。

对于那些来这里寻找另一个常见错误的人，为什么文件上传不起作用但普通表单数据会：确保format="multipart"在创建请求时设置。

这也给出了类似的问题，@RobinElvin在评论中指出

这是因为我缺少 format='multipart'

于 2014-12-20T02:38:15.117 回答
19

Python 3 用户：确保open文件位于mode='rb'(read,binary) 中。否则，当 Django 调用read文件时，utf-8编解码器将立即开始阻塞。该文件应解码为二进制而不是 utf-8、ascii 或任何其他编码。

# This won't work in Python 3
with open(tmp_file.name) as fp:
        response = self.client.post('my_url', 
                                   {'image': fp}, 
                                   format='multipart')

# Set the mode to binary and read so it can be decoded as binary
with open(tmp_file.name, 'rb') as fp:
        response = self.client.post('my_url', 
                                   {'image': fp}, 
                                   format='multipart')
于 2015-05-17T19:23:34.417 回答
9

您可以使用 Django 内置的 SimpleUploadedFile：

from django.core.files.uploadedfile import SimpleUploadedFile

class TestFileUpload(APITestCase):
...

    def test_file_is_accepted(self):
        ...

       tmp_file = SimpleUploadedFile(
                      "file.jpg", "file_content", content_type="image/jpg")

       response = self.client.post(
                      'my_url', {'image': tmp_file}, format='multipart')
       self.assertEqual(response.status_code, status.HTTP_201_CREATED)
于 2019-11-01T13:37:54.173 回答
6

如果要使用 PATCH 方法，要了解如何做到这一点并不是那么简单，但是我在这个问题中找到了解决方案。

from django.test.client import BOUNDARY, MULTIPART_CONTENT, encode_multipart

with open(tmp_file.name, 'rb') as fp:
    response = self.client.patch(
        'my_url', 
        encode_multipart(BOUNDARY, {'image': fp}), 
        content_type=MULTIPART_CONTENT
    )
于 2018-06-01T17:19:15.233 回答
1

对于那些在 Windows 中的人来说，答案有点不同。我必须执行以下操作：

resp = None
with tempfile.NamedTemporaryFile(suffix='.jpg', delete=False) as tmp_file:
    image = Image.new('RGB', (100, 100), "#ddd")
    image.save(tmp_file, format="JPEG")
    tmp_file.close()

# create status update
with open(tmp_file.name, 'rb') as photo:
    resp = self.client.post('/api/articles/', {'title': 'title',
                                               'content': 'content',
                                               'photo': photo,
                                               }, format='multipart')
os.remove(tmp_file.name)

正如这个答案（https://stackoverflow.com/a/23212515/72350）中所指出的那样，该文件在 Windows 中关闭后无法使用。在 Linux 下，@Meistro 的答案应该有效。

于 2017-04-04T15:58:53.693 回答