如何正确使用gophishpythonapi来避免Json的混乱？

Question

因此，我尝试在python3.7项目中使用Gophish，允许用户创建活动、配置文件和所有可以在UI中正常执行的好东西。因此，在查看包文档时，我尝试使用代码创建一个新的活动或用户组或其他任何东西，但不断出现JSON错误。你知道吗

使用python3.7和gophish0.7.1，尝试更改名称，甚至使用直接输入而不是硬编码值，这些都解决不了任何问题，而且我完全不理解Json，我只想让python程序正常工作

def newCampaign():
    groups = [Group(name='testgroup')]
    page = Page(name='Linkedin')
    template = Template(name='Linkedin')
    smtp = SMTP(name='LinkedIN')
    url = 'http://192.168.1.20'
    campaign = Campaign(name='Example Campaign', groups=groups, page=page, template=template, smtp=smtp)
    campaign = api.campaigns.post(campaign)


newCmpgnBut = Button(self, text="NEW CAMPAIGN", command=newCampaign(), bg='black', fg='white')
newCmpgnBut.pack()

我希望这会在我的Gophish创建一个新的活动，相反，我得到的只是一堆乱七八糟的东西：

File "/PycharmProjects/PenTester2.0/phishing.py", line 45, in <module>
    newCmpgnBut = Button(self, text="NEW CAMPAIGN", command=newCampaign(), bg='black', fg='white')
  File "/PycharmProjects/PenTester2.0/phishing.py", line 40, in newCampaign
    campaign = api.campaigns.post(campaign)
  File "/usr/local/lib/python3.7/dist-packages/gophish/api/campaigns.py", line 20, in post
    return super(API, self).post(campaign)
  File "/usr/local/lib/python3.7/dist-packages/gophish/api/api.py", line 92, in post
    raise Error.parse(response.json())
  File "/usr/lib/python3/dist-packages/requests/models.py", line 897, in json
    return complexjson.loads(self.text, **kwargs)
  File "/usr/lib/python3.7/json/__init__.py", line 348, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python3.7/json/decoder.py", line 337, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/usr/lib/python3.7/json/decoder.py", line 355, in raw_decode
    raise JSONDecodeError("Expecting value", s, err.value) from None
json.decoder.JSONDecodeError: Expecting value: line 1 column 1 (char 0)

Process finished with exit code 1

Answer 1

所以在API中/活动.py对于所有其他api文件，第7行的路径中还有一个“/”，这会导致在访问该文件的地址中添加两个斜杠。删除斜杠解决了这个问题