Python: try-except Expression always get the default value when I try to apply it on http request

Question

I use some google API as fellow:

def get_address(lat, lng):
    url = "https://maps.googleapis.com/maps/api/geocode/json?{}".\
      format(urllib.parse.urlencode(args))
    ...
    try:
       r = requests.get(url)
       ...
       return r
    except OSError as e:
       raise NetException(e.message, 400)

while I try use the try-exception handle the excption if the network errors. the try-exception Expression is from here

def try_except(success, failure, *exceptions):
    try:
        return success()
    except exceptions or Exception:
        return failure() if callable(failure) else failure

But when I try to use these exception I always get the failure reults of http, even if I get successful result if I simply run the success function.

>>> re=get_address(-33.865, 151.2094)
>>> re
'Sydney'
>>> r=try_except(get_address(-33.865, 151.2094),"")
>>> r
''

How to make sure the successful result would get correct string reuslt, while the onlly failture of http request get the failure result?

Answer 1

You have to pass the function as success argument. Currently in

r=try_except(get_address(-33.865, 151.2094),"")

you are passing result value of get_address(-33.865, 151.2094) which is 'Sydney' . The actual error is raised on trying to call success() which translates to 'Sydney'() - something like str object is not callable

Proper call would be

r=try_except(lambda: get_address(-33.865, 151.2094), '')