python - 如何在Python中编写可用于异常/错误的if-else语句？

Question

我正在使用Twilio的API返回有关电话号码的信息。一些电话号码无效并返回错误，例如

Traceback (most recent call last):
  File "test_twilio.py", line 17, in <module>
    number = client.lookups.phone_numbers("(4154) 693-
6078").fetch(type="carrier")
  File "/Users/jawnsano/anaconda/lib/python2.7/site-
packages/twilio/rest/lookups/v1/phone_number.py", line 158, in fetch
    params=params,
  File "/Users/jawnsano/anaconda/lib/python2.7/site-
packages/twilio/base/version.py", line 82, in fetch
    raise self.exception(method, uri, response, 'Unable to fetch 
record')
twilio.base.exceptions.TwilioRestException: 
HTTP Error Your request was:

GET /PhoneNumbers/(4154) 693-6078

Twilio returned the following information:

Unable to fetch record: The requested resource /PhoneNumbers/(4154) 
693-6078 was not found

More information may be available here:

https://www.twilio.com/docs/errors/20404

如果返回如上所示的错误，我想打印“有错误”。但是，对于我的if语句，是否有一种方法可以使Python打印出通常存在回溯错误/错误的情况？我认为可能有比设置使其更好的方法

if returned_value = (super long error message):
    etc...

Answer 1

您可以使用try和except来捕获错误。

from twilio.base.exceptions import TwilioRestException

try:
    ... your code
except TwilioRestException:
    print("whatever")