python - 使用Python requests模块上传文件

Question

我需要使用soap端点url加载文件...当我使用下面的代码加载它时，文件正在加载，但它们不是可读的格式...当我使用SOAPUI工具加载时，它会加载正确地...

import requests

xml = '''<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
                xmlns:v1="http://s.sa.com/services/Attachment/v1.0">
   <soapenv:Header/>
   <soapenv:Body>
      <v1:attachment>
         <filename>FUZZY.csv</filename>
         <data>cid:138641430598</data>
      </v1:attachment>
   </soapenv:Body>
</soapenv:Envelope>'''
target_url =  'https://s.sa.com:443/soatest/FileAttachmentService'
headers = {'Content-Type': 'text/xml','charset':'utf-8'}
r = requests.post(target_url,data=xml,headers=headers,auth=('3user1',''))
print 'r.text = ', r.text
print 'r.content = ', r.content
print 'r.status_code = ', r.status_code

新变化:-

files = {'file':open('./FUZZY.csv','rb')}
print files
r = requests.post(target_url,files=files,data=xml,headers=headers,auth=('p3user1',''))

错误:

Traceback (most recent call last):
  File "soapcall_python.py", line 18, in <module>
    r = requests.post(target_url,files=files,data=xml,headers=headers,auth=('p3user1',''))
  File "/opt/python2.7/lib/python2.7/site-packages/requests-2.3.0-py2.7.egg/requests/api.py", line 88, in post
    return request('post', url, data=data, **kwargs)
  File "/opt/python2.7/lib/python2.7/site-packages/requests-2.3.0-py2.7.egg/requests/api.py", line 44, in request
    return session.request(method=method, url=url, **kwargs)
  File "/opt/python2.7/lib/python2.7/site-packages/requests-2.3.0-py2.7.egg/requests/sessions.py", line 418, in request
    prep = self.prepare_request(req)
  File "/opt/python2.7/lib/python2.7/site-packages/requests-2.3.0-py2.7.egg/requests/sessions.py", line 356, in prepare_request
    hooks=merge_hooks(request.hooks, self.hooks),
  File "/opt/python2.7/lib/python2.7/site-packages/requests-2.3.0-py2.7.egg/requests/models.py", line 297, in prepare
    self.prepare_body(data, files)
  File "/opt/python2.7/lib/python2.7/site-packages/requests-2.3.0-py2.7.egg/requests/models.py", line 432, in prepare_body
    (body, content_type) = self._encode_files(files, data)
  File "/opt/python2.7/lib/python2.7/site-packages/requests-2.3.0-py2.7.egg/requests/models.py", line 109, in _encode_files
    raise ValueError("Data must not be a string.")
ValueError: Data must not be a string.

Answer 1

您没有将文件的内容发送到任何地方。您只是发送对服务器可以看到的任何地方都不存在的文件的引用。

作为 SOAP references to attachments 的文档解释说，执行此操作的方法是发送 MIME 多部分消息。如果您使用 CID 引用机制，则 cid 不是任意字符串，它必须与 MIME 信封中邮件的 Content-ID header 相匹配。

POST a Multipart-Encoded File 的请求文档解释如何在 MIME 请求中将文件内容作为消息发送；简而言之:

with open('FUZZY.csv', 'rb') as f:
    files = {'file': f}
    r = requests.post(target_url,
                      data=xml, headers=headers, auth=('3user1',''), 
                      files=files)

但是，这种简单的方法无法让您访问将在后台为您的消息生成的 Content-ID。因此，如果您想使用 CID 引用机制，则需要手动生成 MIME 信封(例如，通过使用 email.mail.MIMEMultipart )并将整个内容作为 data 字符串发送。