python请求：如何获取“异常代码”

Question

I am using "requests (2.5.1)" .now I want to catch the exception and return an dict with some exception message,the dict I will return is as following:

{
   "status_code":   61, # exception code,
   "msg": "error msg",
}

but now I can't get the error status_code and error message,I try to use

try:
.....
except requests.exceptions.ConnectionError as e:
response={
            u'status_code':4040,
            u'errno': e.errno,
            u'message': (e.message.reason),
            u'strerror': e.strerror,
            u'response':e.response,
        }

but it's too redundancy,how can I get the error message simplicity?anyone can give some idea?

Answer 1

try:
    #the codes that you want to catch errors goes here
except:
    print ("an error occured.")

This going to catch all errors, but better you define the errors instead of catching all of them and printing special sentences for errors.Like;

try:
    #the codes that you want to catch errors goes here
except SyntaxError:
    print ("SyntaxError occured.")
except ValueError:
    print ("ValueError occured.")
except:
    print ("another error occured.")

Or;

try:
    #the codes that you want to catch errors goes here 
except Exception as t:
    print ("{} occured.".format(t))