使用Django的测试客户端,使用内存文件发出请求

Question

我正在为Django视图编写测试,我想发布一个文件.这是一个相当简单的测试,我不想tests/用不同的文本文件丢弃我的目录,所以我想使用内存文件并动态创建内容:

from StringIO import StringIO
file = StringIO('content')
client.post("/", data={'file': file})

不幸的是,这不起作用:

Traceback (most recent call last):
  File "/Users/brad/project/tests/files.py", line 57, in test_set_and_save
    'mgmt-current_step': 'Attachments',
  File "/Users/brad/django/test/client.py", line 423, in post
    response = super(Client, self).post(path, data=data, content_type=content_type, **extra)
  File "/Users/brad/django/test/client.py", line 245, in post
    post_data = self._encode_data(data, content_type)
  File "/Users/brad/django/test/client.py", line 211, in _encode_data
    return encode_multipart(BOUNDARY, data)
  File "/Users/brad/django/test/client.py", line 117, in encode_multipart
    lines.extend(encode_file(boundary, key, value))
  File "/Users/brad/django/test/client.py", line 145, in encode_file
    content_type = mimetypes.guess_type(file.name)[0]
AttributeError: StringIO instance has no attribute 'name'

Answer 1

Django附带了一组Python内置file对象的包装器.在这种情况下django.core.files.base.ContentFile是合适的:

from django.core.files.base import ContentFile
file = ContentFile(b'content', name='plain.txt')
client.post('/', data={'file': file})

ContentFile 期望使用字节,所以不要给它unicode数据.

另一个技巧(如果你不关心文件的内容)是发送当前文件:

client.post('/', data={'file': open(__file__, 'rb'))