如何在zeep中访问该方法

Question

我正在尝试下面的代码，但我得到zeep.exceptions.Fault：null错误

from zeep import Client
from zeep.transports import Transport
from requests_ntlm import HttpNtlmAuth
from requests import Session
session = Session()
username = ""
password = ""
url = ""
session.auth = HttpNtlmAuth(username,password)
transport = Transport(session=session)
client = Client(url,transport=transport)
client.service.CI_Z_SP_PERS_DTL_CI_G(EMPLID="001670")

Answer 1

这不是答案，但以下方法可以帮助捕获代码中的错误源。要在代码中获取导致line / snippet的实际错误，可以使用try / except块并捕获错误：

#<---code--->
try:
    session.auth = HttpNtlmAuth(username,password)
    transport = Transport(session=session)
    client = Client(url,transport=transport)
    client.service.CI_Z_SP_PERS_DTL_CI_G(EMPLID="001670")
except zeep.exceptions.Fault as fault:
    print fault.message
    print fault.code
    print fault.actor

要么

except zeep.exceptions.Fault as fault:
    print fault.__dict__

这将有助于调试错误发生的位置。