[Tutor] Interactive Menu Woes

Bryan Magalski bryan_magalski at yahoo.com
Wed Nov 14 21:40:04 CET 2007


Thank you for your suggestion.  I did not create the original script, so it will stay as is and my addition for the menu has been adjusted.  

Now that I can make a clear distinction of what I am returning, I am getting a new error that leads me that I am trying to call a function that cannot be seen when instantiated:

To add a name, please input he following information: 
Name: Bryan
Number: 1234567890
What type of number is this? (choose one): 
                     1. Home:
                     2. Work:
                     3. Cell:
                     : 1
Traceback (most recent call last):
  File "menu_modified.py", line 95, in <module>
    menu.addName()
  File "menu_modified.py", line 73, in addName
    enter(name, number, returnType)
AttributeError: phoneentry instance has no __call__ method

[code]
class MenuInput:

        # ask user to input data and store to be passed to manageable variables.
        def addName(self):
                print "To add a name, please input he following information: "
                name = raw_input("Name: ")
                number = raw_input("Number: ")
                get = int(raw_input("What type of number is this? (choose one): \n \
                    1. Home:\n \
                    2. Work:\n \
                    3. Cell:\n \
                    : "))
                returnType = self.numberType(get)
                               
                #print returnType
                enter = phoneentry()
                enter(name, number, returnType)
[/code]

I googled the __call__ method, but I am not getting to the meat of what all of the leads have to say.  I know it has to deal with when I instantiated phoneentry() within addName(), I just don't know how to fix it or "call" it properly.

Again, thanks in advance.

----- Original Message ----
From: Evert Rol <evert.rol at gmail.com>
To: Bryan Magalski <bryan_magalski at yahoo.com>
Cc: tutor at python.org
Sent: Wednesday, November 14, 2007 1:28:45 PM
Subject: Re: [Tutor] Interactive Menu Woes


> I am trying to build a menu for the following script to make it  
> more "user friendly".  Nothing fancy, just a simple add data and  
> look up the entered results.
>
> The problem is that when I run my modified version with this  
> snippet (please see attachment for original and my modified
 versions):
>
> [code]
> class MenuInput:
>
>  # ask user to input data and store to be passed to manageable  
> variables.
>         def addName(self):
>                 print "To add a name, please input he following  
> information: "
>                 self.enter = phoneentry()
>                 self.name = raw_input("Name: ")
>                 self.number = raw_input("Number: ")
>                 self.get = int(raw_input("What type of number is  
> this? (choose one): \n 1. Home:\n 2. Work:\n 3. Cell:\n : "))
>                 def
 numberType(self, get = 1):
>                         if self.get == 1:
>                                 self.returnType = HOME
>                                 #return self.getType
>                         elif self.gete == 2:
>                                 self.returnType = WORK
>                                 #return self.getType
>                         elif self.get == 3:
>       
                          self.returnType = FAX
>                                 #return self.getType
>                         return self.returnType
>
>                 self.type = numberType(self.get)
>                 self.enter(self.name, self.number, self.returnType)
>


Bypassing the whole 'function within function' problem, consider  
using a dictionary instead: somedict = {1: HOME, 2: WORK, 3: FAX},  
and self.type = somedict[self.get]. This also avoids endless if-elif- 
elif-elif sequences (looking at showtype() in your original script,  
where you can
 apply the same technique); it's often clearer, and  
certainly shorter.










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