Type error in python program

Question

I always get a Type Error when I run the following python code (abc.py) as follows:

./abc.py activatelink alphabeta
Type Error: ['alphabeta']

My code:

#!/usr/bin/python

import urllib2
from urllib2 import URLError
from urllib2 import HTTPError
import requests
import urllib
import json
import time
import os
import sys
import hashlib


def activate_user(link):
        print invoke_rest('GET', link)

def invoke_rest(request_type, rest_url, payload, headers):
        try:
                api_url = rest_url
                if request_type == 'GET':
                        r = requests.get(api_url)
                        to_ret = {'code':r.status_code, 'reply':r.text}
                        return to_ret
                elif request_type == 'POST':
                        r = requests.post(api_url, data=payload, headers=headers)
                        to_ret = {'code':r.status_code, 'reply':r.text}
                        return to_ret
                else:
                        return "Invalid request type ", request_type
        except Exception, e:
                return "Exception:", e, " in getting the API call"

def help():
    print ('Usage: %s { activate | help }', os.path.basename(sys.argv[0])

if __name__ == '__main__':
    actions = {'activatelink': activate_user, 'help': help}
    try:
        action = str(sys.argv[1])
    except IndexError:
        print "IndexError: ", sys.argv[1]
        action = 'help'
    args = sys.argv[2:]
    try:
        actions[action](*args)
    except (KeyError):
        print "Key Error:", args
        help()
    except (TypeError):
        print "Type Error:", args
        help()

Am I doing anything wrong? I added some other functions other than activatelink, which work fine, can anyone point out whats wrong in here?

Can you remove the exception handlers in your code so we can see the proper full exception with traceback rather than the print message? — Martijn Pieters
– Martijn Pieters, Commented Jun 16, 2015 at 12:25

Martijn Pieters · Accepted Answer · 2015-06-16 12:27:59Z

6

Your invoke_rest() function takes four arguments:

def invoke_rest(request_type, rest_url, payload, headers):

but you pass in just the two:

print invoke_rest('GET', link)

That raises a TypeError exception:

>>> def invoke_rest(request_type, rest_url, payload, headers):
...     pass
... 
>>> invoke_rest('GET', 'alphabeta')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: invoke_rest() takes exactly 4 arguments (2 given)

Perhaps you wanted those two extra arguments (payload and headers) to be optional. If so, make them keyword arguments and set their default value to None:

def invoke_rest(request_type, rest_url, payload=None, headers=None):

which is fine by the requests library.

answered Jun 16, 2015 at 12:27

Martijn Pieters

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