我正处于将我们的网络应用程序与 PayPal 的快速结帐 API 集成的第一阶段。要进行购买,我当然必须使用我们的客户 ID 和客户密码获取 Bearer token 。
我使用以下 curl 命令成功获取该 token :
curl https://api.sandbox.paypal.com/v1/oauth2/token \
-H "Accept: application/json" \
-H "Accept-Language: en_US" \
-u "ourID:ourSecret" \
-d "grant_type=client_credentials"
现在我正尝试使用 urllib2 在 python 中实现相同的结果。我得到了以下代码,它产生了 401 HTTP Unauthorized 异常。
import urllib
import urllib2
url = "https://api.sandbox.paypal.com/v1/oauth2/token"
PAYPAL_CLIENT_ID = "ourID"
PAYPAL_CLIENT_SECRET = "ourSecret"
passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, url, PAYPAL_CLIENT_ID, PAYPAL_CLIENT_SECRET)
authhandler = urllib2.HTTPBasicAuthHandler(passman)
opener = urllib2.build_opener(authhandler)
urllib2.install_opener(opener)
req = urllib2.Request( url=url,
headers={
"Accept": "application/json",
"Accept-Language": "en_US",
},
data =urllib.urlencode({
"grant_type":"client_credentials",
}),)
result = urllib2.urlopen(req).read()
print result
有人知道我上面做错了什么吗?非常感谢您的任何见解
请您参考如下方法:
遇到同样的问题。基于Get access token from Paypal in Python - Using urllib2 or requests library工作的 python 代码是:
import urllib
import urllib2
import base64
token_url = 'https://api.sandbox.paypal.com/v1/oauth2/token'
client_id = '.....'
client_secret = '....'
credentials = "%s:%s" % (client_id, client_secret)
encode_credential = base64.b64encode(credentials.encode('utf-8')).decode('utf-8').replace("\n", "")
header_params = {
"Authorization": ("Basic %s" % encode_credential),
"Content-Type": "application/x-www-form-urlencoded",
"Accept": "application/json"
}
param = {
'grant_type': 'client_credentials',
}
data = urllib.urlencode(param)
request = urllib2.Request(token_url, data, header_params)
response = urllib2.urlopen(request).open()
print response
我相信原因在 Python urllib2 Basic Auth Problem 中有解释。
Python libraries, per HTTP-Standard, first send an unauthenticated request, and then only if it's answered with a 401 retry, are the correct credentials sent. If the servers don't do "totally standard authentication" then the libraries won't work.