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python - Django Sendgrid 身份验证错误

转载 作者:太空宇宙 更新时间:2023-11-04 03:38:26 25 4
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我正在尝试通过 SendGrid API 对自己进行身份验证,以便使用模板引擎 API(为 A/B 测试创建它们),但无论如何,我都会收到 401 未经授权的错误。有人可以帮忙吗?难道我做错了什么 ?这是我的代码:

@receiver(models.signals.post_save, sender=Template)
def create_sendgrid_template(sender, instance, created, *args, **kwargs):
url = 'https://api.sendgrid.com/v3/templates'
sendgrid_data = {
"name": "example_template",
}
if created:
request = urllib2.Request(url)
base64string = base64.encodestring(
'%s:%s' % (
SENDGRID_API_USER,
SENDGRID_API_PASSWORD)
).replace('\n', '')
request.add_header("Authorization", "Basic %s" % base64string)
result = urllib2.urlopen(request)

最佳答案

尝试对您的数据进行编码并将其添加到请求中——我怀疑您正在获取而不是发布。查看urllib howto .还值得验证您的凭据是否正确。

@receiver(models.signals.post_save, sender=Template)
def create_sendgrid_template(sender, instance, created, *args, **kwargs):
url = 'https://api.sendgrid.com/v3/templates'
sendgrid_data = {
"name": "example_template",
}
data = json.dumps(sendgrid_data)

if created:
request = urllib2.Request(url, data)
base64string = base64.encodestring(
'%s:%s' % (
SENDGRID_API_USER,
SENDGRID_API_PASSWORD)
).replace('\n', '')
request.add_header("Authorization", "Basic %s" % base64string)
request.add_header('Content-type', 'application/json')
result = urllib2.urlopen(request)

如果您不想推出自己的解决方案,可以使用 sendwithus 等服务和 mailjet对电子邮件进行 A/B 测试。

关于python - Django Sendgrid 身份验证错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27769606/

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