python 请求发送数组中的文件显示 ValueError : too many values to unpack

Question

这里有什么问题吗？

我收到错误显示

ValueError:太多值无法解压

此代码过程是获取文件夹中的所有可用图像，然后将该图像位置放入数组中。然后创建另一个名为files的数组并添加特定格式的图像位置并向API发送POST请求

import requests
import logging
import os
import json

try:
    import http.client as http_client
except ImportError:
    # Python 2
    import httplib as http_client
http_client.HTTPConnection.debuglevel = 1

logging.basicConfig()
logging.getLogger().setLevel(logging.DEBUG)
requests_log = logging.getLogger("requests.packages.urllib3")
requests_log.setLevel(logging.DEBUG)
requests_log.propagate = True

url_and = "https://api.test.com/api/3.0/listings/"

android_token = '68as76df87s86df7asd76f87as6df78sfd'
headers = {
    'Authorization': "Token " + android_token,
    'platform': 'android'
}

data_android = {
    'mailing_details':'3',
    'abcoupay':'false',
    'price':'55.00',
    'description':'Test description',
    'title':'TEST drill machine and others.',
    'meetup':'false',
    'condition':'2',
    'mailing':'true',
    'collection_id':'24'
}

urls = []
for file in os.listdir(os.getcwd()+"/product_images"):
    if file.endswith((".jpg",".jpeg",".png",".JPG",".JPEG",".PNG")):
        x = os.getcwd()+"\\"+file
        urls.append(x)

files = []
x = 0
for file in urls:
    files.append("'photo_"+str(x)+"': ('image_"+str(x)+".jpg', open('"+file+"', 'rb'), 'image/jpeg')")
    x+=1

# files = {
#             'photo_0': ('image_0.jpg', open('E:/products files/Drill machine/1.jpg', 'rb'), 'image/jpeg'),
#             'photo_1': ('image_1.jpg', open('E:/products files/Drill machine/2.jpg','rb'), 'image/jpeg')        
#         }

response = requests.request("POST", url_and,data=data_android,files=files,headers=headers)
print(response.text.encode("utf-8"))

错误

Traceback (most recent call last):
  File "cookies.py", line 102, in <module>
    response = requests.request("POST", url_and,data=data_android,files=files,headers=headers)
  File "C:\python27\lib\site-packages\requests\api.py", line 60, in request
    return session.request(method=method, url=url, **kwargs)
  File "C:\python27\lib\site-packages\requests\sessions.py", line 519, in request
    prep = self.prepare_request(req)
  File "C:\python27\lib\site-packages\requests\sessions.py", line 462, in prepare_request
    hooks=merge_hooks(request.hooks, self.hooks),
  File "C:\python27\lib\site-packages\requests\models.py", line 316, in prepare
    self.prepare_body(data, files, json)
  File "C:\python27\lib\site-packages\requests\models.py", line 504, in prepare_body
    (body, content_type) = self._encode_files(files, data)
  File "C:\python27\lib\site-packages\requests\models.py", line 141, in _encode_files
    for (k, v) in files:
ValueError: too many values to unpack

Answer 1

requests.request的files参数应该是一个字典，其中键是文件名，值是文件内容或文件对象。因此，您应该将 files 变量构建为字典:

files = {'image_%s.jpg' % x: open(file, 'rb') for x, file in enumerate(urls)}

如果您还需要内容类型，您可以将字典值设置为包含文件名、文件对象和内容类型的三元组:

files = {'image_%s.jpg' % x: ('image_%s.jpg' % x, open(file, 'rb'), 'image/jpeg') for x, file in enumerate(urls)}